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Edexcel GCSE Chemistry

Topic questions Quantitative chemistry Percentage yield

Percentage yield

15 questions · 163 marks · about 3 hours

Percentage yield questions

Edexcel 1CH0 Percentage yield 15 questions

Paper 1 June 2025 Foundation tier, Question 10, also on the Higher paper 11 marks

Phosphoric acid is important in the production of fertilisers.

(a)
(a)(i)

A few drops of litmus solution are added to a sample of dilute phosphoric acid.

What is the colour of the mixture formed?

Question 10, shown as an image
[1]
Reveal mark scheme Mark scheme for question 10(a)(i)
(a)(ii)

State why litmus cannot be used to measure pH.

[1]
Reveal mark scheme Mark scheme for question 10(a)(ii)
(b)

Phosphoric acid can react to form phosphate salts which are used in fertilisers.

Give two reasons why farmers use fertilisers on their fields.

[2]
Reveal mark scheme Mark scheme for question 10(b)
(c)

An experiment was carried out to produce phosphoric acid.

(c)(i)

The actual yield for the experiment was 18.62 g.

The theoretical yield for the experiment is 24.50 g.

Question 10, shown as an image

Calculate the percentage yield of this experiment.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 10(c)(i)
(c)(ii)

The equation for a reaction to make phosphoric acid, H3PO4, is

Question 10, shown as an image

Calculate the atom economy for the production of phosphoric acid in this reaction.

Question 10, shown as an image

(relative formula masses: Ca5(PO4)3OH = 502, H2SO4 = 98, H3PO4 = 98, CaSO4 = 136, H2O = 18)

atom economy =........ %

[3]
Reveal mark scheme Mark scheme for question 10(c)(ii)
(c)(iii)

The equation for a reaction to produce phosphoric acid is

Question 10, shown as an image

Explain how the atom economy of the reaction can be increased.

[2]
Reveal mark scheme Mark scheme for question 10(c)(iii)

Paper 1 June 2025 Higher tier, Question 4, also on the Foundation paper 11 marks

Also involves NPK fertilisers

Phosphoric acid is important in the production of fertilisers.

(a)
(a)(i)

A few drops of litmus solution are added to a sample of dilute phosphoric acid.

What is the colour of the mixture formed?

Question 4, shown as an image
[1]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

State why litmus cannot be used to measure pH.

[1]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(b)

Phosphoric acid can react to form phosphate salts which are used in fertilisers.

Give two reasons why farmers use fertilisers on their fields.

[2]
Reveal mark scheme Mark scheme for question 4(b)
(c)

An experiment was carried out to produce phosphoric acid.

(c)(i)

The actual yield for the experiment was 18.62 g.

The theoretical yield for the experiment is 24.50 g.

Question 4, shown as an image

Calculate the percentage yield of this experiment.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 4(c)(i)
(c)(ii)

The equation for a reaction to make phosphoric acid, H3PO4, is

Question 4, shown as an image

Calculate the atom economy for the production of phosphoric acid in this reaction.

Question 4, shown as an image

(relative formula masses: Ca5(PO4)3OH = 502, H2SO4 = 98, H3PO4 = 98, CaSO4 = 136, H2O=18)

atom economy =........

[3]
Reveal mark scheme Mark scheme for question 4(c)(ii)
(c)(iii)

The equation for a reaction to produce phosphoric acid is

Question 4, shown as an image

Explain how the atom economy of the reaction can be increased.

[2]
Reveal mark scheme Mark scheme for question 4(c)(iii)

Paper 1 June 2024 Foundation tier, Question 10 11 marks

(a)

Titanium can be extracted from titanium oxide, TiO2, by reaction with magnesium.

(a)(i)

100 tonnes of titanium oxide was heated with magnesium. The titanium formed in the reaction was separated and purified. The mass of titanium was then determined.

The results are shown in Figure 16.

Figure for question 10

Use the information in Figure 16 to calculate the percentage yield of titanium in this process.

Question 10, shown as an image

Give your answer to 1 decimal place.

percentage yield =........

[3]
Reveal mark scheme Mark scheme for question 10(a)(i)
(a)(ii)

Give two reasons why the percentage yield for this process is less than 100%.

[2]
Reveal mark scheme Mark scheme for question 10(a)(ii)
(a)(iii)

The balanced equation for this process is

Question 10, shown as an image

Calculate the atom economy of this process to produce titanium.

Question 10, shown as an image

Give your answer to 2 significant figures.

(relative atomic masses: O = 16, Mg = 24, Ti = 48)

atom economy =........

[3]
Reveal mark scheme Mark scheme for question 10(a)(iii)
(b)

Ethanol, C2H5OH, can be produced by two different methods.

β€’ by the hydration of ethene, C2H4

Question 10, shown as an image

β€’ and by the fermentation of a carbohydrate, e.g. sucrose, C12H22O11

Question 10, shown as an image
(b)(i)

State why the hydration of ethene has an atom economy of 100%.

[1]
Reveal mark scheme Mark scheme for question 10(b)(i)
(b)(ii)

Explain how the atom economy of the fermentation reaction can be improved.

[2]
Reveal mark scheme Mark scheme for question 10(b)(ii)

Paper 1 June 2024 Higher tier, Question 4 11 marks

Also involves Producing ethanol

(a)

Titanium can be extracted from titanium oxide, TiO2, by reaction with magnesium.

(a)(i)

100 tonnes of titanium oxide was heated with magnesium. The titanium formed in the reaction was separated and purified. The mass of titanium was then determined.

The results are shown in Figure 4.

Figure for question 4

Use the information in Figure 4 to calculate the percentage yield of titanium in this process.

Question 4, shown as an image

Give your answer to 1 decimal place.

percentage yield =........

[3]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

Give two reasons why the percentage yield for this process is less than 100%.

[2]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(a)(iii)

The balanced equation for this process is

Question 4, shown as an image

Calculate the atom economy of this process to produce titanium.

Question 4, shown as an image

Give your answer to 2 significant figures.

(relative atomic masses: O = 16, Mg = 24, Ti = 48)

atom economy =........

[3]
Reveal mark scheme Mark scheme for question 4(a)(iii)
(b)

Ethanol, C2H5OH, can be produced by two different methods.

β€’ by the hydration of ethene, C2H4

Question 4, shown as an image

β€’ and by the fermentation of a carbohydrate, e.g. sucrose, C12H22O11

Question 4, shown as an image
(b)(i)

State why the hydration of ethene has an atom economy of 100%.

[1]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

Explain how the atom economy of the fermentation reaction can be improved.

[2]
Reveal mark scheme Mark scheme for question 4(b)(ii)

Paper 1 June 2023 Foundation tier, Question 10 11 marks

There are several stages to the production of sulfuric acid in industry.

(a)

Sulfur dioxide is required for the production of sulfuric acid.

Sulfur dioxide can be obtained by heating copper sulfide, Cu2S, in excess air.

Question 10, shown as an image

Calculate the atom economy for the production of sulfur dioxide, SO2, in this reaction.

(relative atomic mass: Cu = 63.5 relative formula masses: O2 = 32.0, Cu2S = 159.0, SO2 = 64.0)

Give your answer to two significant figures.

[4]
Reveal mark scheme Mark scheme for question 10(a)
(b)

In one stage vanadium oxide, V2O5, is used.

Based on the position of vanadium in the periodic table, which row shows the most likely melting point of vanadium and colour of vanadium oxide?

Figure for question 10
[1]
Reveal mark scheme Mark scheme for question 10(b)
(c)

The equation shows a reaction forming sulfuric acid.

(c)(i)

Calculate the maximum mass of sulfuric acid that could be produced from 400 tonnes of sulfur trioxide, SO3.

(relative formula masses: SO3 = 80, H2SO4 = 98)

maximum mass of sulfuric acid =........ tonnes

[2]
Reveal mark scheme Mark scheme for question 10(c)(i)
(c)(ii)

Using a different amount of sulfur trioxide, it was calculated that 700 tonnes of sulfuric acid could be made.

The actual mass produced was 672 tonnes.

Calculate the percentage yield of sulfuric acid.

[2]
Reveal mark scheme Mark scheme for question 10(c)(ii)
(c)(iii)
Figure for question 10 Figure for question 10
[2]
Reveal mark scheme Mark scheme for question 10(c)(iii)

Paper 1 June 2023 Higher tier, Question 4 11 marks

There are several stages to the production of sulfuric acid in industry.

(a)

Sulfur dioxide is required for the production of sulfuric acid.

Sulfur dioxide can be obtained by heating copper sulfide, Cu2S, in excess air.

Question 4, shown as an image

Calculate the atom economy for the production of sulfur dioxide, SO2, in this reaction.

(relative atomic mass: Cu = 63.5 relative formula masses: O2 = 32.0, Cu2S = 159.0, SO2 = 64.0)

Give your answer to two significant figures.

atom economy =........

[4]
Reveal mark scheme Mark scheme for question 4(a)
(b)

In one stage vanadium oxide, V2O5, is used.

Based on the position of vanadium in the periodic table, which row shows the most likely melting point of vanadium and colour of vanadium oxide?

Figure for question 4
[1]
Reveal mark scheme Mark scheme for question 4(b)
(c)

The equation shows a reaction forming sulfuric acid.

(c)(i)

Calculate the maximum mass of sulfuric acid that could be produced from 400 tonnes of sulfur trioxide, SO3.

(relative formula masses: SO3 = 80, H2SO4 = 98)

maximum mass of sulfuric acid =........ tonnes

[2]
Reveal mark scheme Mark scheme for question 4(c)(i)
(c)(ii)

Using a different amount of sulfur trioxide, it was calculated that 700 tonnes of sulfuric acid could be made.

The actual mass produced was 672 tonnes.

Calculate the percentage yield of sulfuric acid.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 4(c)(ii)
(c)(iii)

State two reasons why the percentage yield is less than 100 %.

[2]
Reveal mark scheme Mark scheme for question 4(c)(iii)

Paper 1 June 2022 Foundation tier, Question 10 11 marks

Sucrose is a carbohydrate.

When a solution of sucrose is fermented using yeast, ethanol is formed.

sucrose + water maths symbol ethanol + carbon dioxide

(a)

In one experiment, 100.00 g of sucrose was dissolved in water.

Yeast was added and the mixture allowed to ferment until no more bubbles of carbon dioxide were seen to be formed.

The ethanol was obtained from the mixture and its mass determined.

The results are shown in Figure 16.

Figure for question 10

The percentage yield is calculated using

Question 10, shown as an image
(a)(i)

State the meanings of the terms actual yield and theoretical yield.

actual yield

theoretical yield

[2]
Reveal mark scheme Mark scheme for question 10(a)(i)
(a)(ii)
Figure for question 10 Figure for question 10
[2]
Reveal mark scheme Mark scheme for question 10(a)(ii)
(a)(iii)

State two reasons why the actual yield of a reaction is usually less than the theoretical yield.

[2]
Reveal mark scheme Mark scheme for question 10(a)(iii)
(b)

The balanced equation for the fermentation of sucrose is

Question 10, shown as an image
(b)(i)

Calculate the atom economy of this reaction to produce ethanol.

Give your answer to two significant figures.

(relative formula masses: C12H22O11 = 342, H2O = 18, C2H5OH = 46, CO2 = 44)

[3]
Reveal mark scheme Mark scheme for question 10(b)(i)
(b)(ii)

Explain the effect on the atom economy of this reaction if the carbon dioxide produced was used to make fizzy drinks.

[2]
Reveal mark scheme Mark scheme for question 10(b)(ii)

Paper 1 June 2022 Higher tier, Question 4 11 marks

Also involves Amounts of substance in equations

Sucrose is a carbohydrate.

When a solution of sucrose is fermented using yeast, ethanol is formed.

sucrose + water maths symbol ethanol + carbon dioxide

(a)

In one experiment, 100.00 g of sucrose was dissolved in water.

Yeast was added and the mixture allowed to ferment until no more bubbles of carbon dioxide were seen to be formed.

The ethanol was obtained from the mixture and its mass determined.

The results are shown in Figure 3.

Figure for question 4

The percentage yield is calculated using

Question 4, shown as an image
(a)(i)

State the meanings of the terms actual yield and theoretical yield.

actual yield

[2]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

theoretical yield

Use the information in Figure 3 to calculate the percentage yield of ethanol in this experiment.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(a)(iii)

State two reasons why the actual yield of a reaction is usually less than the theoretical yield.

[2]
Reveal mark scheme Mark scheme for question 4(a)(iii)
(b)

The balanced equation for the fermentation of sucrose is

Question 4, shown as an image
(b)(i)

Calculate the atom economy of this reaction to produce ethanol.

Give your answer to two significant figures.

(relative formula masses: C12H22O11 = 342, H2O = 18, C2H5OH = 46, CO2 = 44)

atom economy =........

[3]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

Explain the effect on the atom economy of this reaction if the carbon dioxide produced was used to make fizzy drinks.

[2]
Reveal mark scheme Mark scheme for question 4(b)(ii)

Paper 1 November 2021 Foundation tier, Question 7 13 marks

Also involves Hazard symbols and risk assessment

(a)

This question is about the metal gold.

(a)(i)

Gold can be hammered into shape.

State the name of this property.

[1]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

Gold alloys can be used to repair teeth. One reason that gold alloys are used is that they can be hammered into shape.

Give one other reason why gold alloys are used to repair teeth.

[1]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(b)

A gold alloy contains 78 % gold by mass.

Calculate the mass of gold in 2.00 kg of this alloy. Give your answer in grams.

[3]
Reveal mark scheme Mark scheme for question 7(b)
(c)

A substance used to purify gold is kept in a container. There are some hazard symbols on the container.

Draw one straight line from each hazard symbol to its meaning.

Figure for question 7
[2]
Reveal mark scheme Mark scheme for question 7(c)
(d)

*(d) Some gold alloys contain copper.

Copper reacts with oxygen when heated.

copper + oxygen maths symbol copper oxide

A teacher calculates that 1.20 g of copper reacts completely with oxygen to form 1.50 g of copper oxide.

A student heats 1.20 g of copper pieces in a container. Then they heat 1.20 g of copper powder in another container. After heating, the mass of the solid in the containers is found.

The results are shown in Figure 7.

Figure for question 7

Explain the observations and give reasons why the masses after heating are less than expected.

[6]
Reveal mark scheme Mark scheme for question 7(d)

Paper 1 November 2021 Higher tier, Question 1 7 marks

Also involves Thermal decomposition

When heated, zinc carbonate decomposes to form zinc oxide and carbon dioxide gas.

Question 1, shown as an image
(a)

A student investigated the decomposition of a sample of zinc carbonate.

The student used the following method.

step 1 the mass of an empty crucible was determined

step 2 a sample of zinc carbonate was placed into the crucible

step 3 the mass of the crucible and the zinc carbonate was determined

step 4 the crucible and zinc carbonate was heated for five minutes

step 5 the mass of the crucible and contents was determined.

Figure 1 shows the apparatus used.

Figure for question 1

Suggest how the student could confirm that the decomposition was complete.

[3]
Reveal mark scheme Mark scheme for question 1(a)
(b)

The theoretical maximum yield of zinc oxide was 1.86 g.

The actual yield was 1.63 g.

Calculate the percentage yield of zinc oxide.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 1(b)
(c)

Another student carried out the experiment using a similar method. The student used a blue, roaring Bunsen burner flame and placed a lid on the crucible.

State why the student used a blue, roaring flame and a lid.

why a blue, roaring flame........

[2]
Reveal mark scheme Mark scheme for question 1(c)

Paper 1 November 2020 Foundation tier, Question 10 11 marks

Potassium hydroxide reacts with hydrochloric acid to form potassium chloride and water.

potassium hydroxide + hydrochloric acid maths symbol potassium chloride + water

(a)

A student carried out a titration to find the exact volume of dilute hydrochloric acid that reacted with 25.0 cm3 of potassium hydroxide solution.

There were five steps in the titration. The steps shown are not in the correct order.

step J pour the potassium hydroxide solution into a conical flask and add a few drops of indicator to this solution step K fill a burette with the dilute hydrochloric acid and record the initial reading from the burette step L use a measuring cylinder to obtain 25 cm3 of potassium hydroxide solution step M take a final reading from the burette and calculate the volume of the dilute hydrochloric acid reacted step N run the dilute hydrochloric acid from the burette into the conical flask until the indicator changes colour

(a)(i)

Write the steps in the correct order.

Some of the steps have been completed for you.

first step last step

Figure for question 10
[1]
Reveal mark scheme Mark scheme for question 10(a)(i)
(a)(ii)

Suggest an alternative piece of apparatus that could be used in step L to obtain exactly 25.0 cm3 of potassium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 10(a)(ii)
(b)

A student was then asked to produce a pure sample of solid potassium chloride.

After finding the volume of acid reacted in step M, the student added this volume of acid to a fresh 25.0 cm3 sample of the potassium hydroxide solution. This mixture was then evaporated.

(b)(i)

Explain why this new mixture was evaporated rather than the original mixture from the titration, to produce a pure sample of solid potassium chloride.

[2]
Reveal mark scheme Mark scheme for question 10(b)(i)
(b)(ii)

After evaporation, the mass of the potassium chloride was determined.

The theoretical yield of the experiment was 0.70 g. The actual yield was 0.84 g.

This gave a percentage yield greater than 100%.

Calculate the percentage yield of this experiment.

percentage yield =........

[2]
Reveal mark scheme Mark scheme for question 10(b)(ii)
(b)(iii)

Suggest a reason why the actual yield was greater than the theoretical yield.

[1]
Reveal mark scheme Mark scheme for question 10(b)(iii)
(b)(iv)

The equation for the reaction between potassium hydroxide solution and dilute hydrochloric acid is

Question 10, shown as an image

Calculate the atom economy for the production of potassium chloride from potassium hydroxide and hydrochloric acid. (relative formula masses: KOH = 56.0, HCl = 36.5, KCl = 74.5, H2O = 18.0)

Give your answer to one decimal place.

atom economy =........

[4]
Reveal mark scheme Mark scheme for question 10(b)(iv)

Paper 1 November 2020 Higher tier, Question 4 11 marks

Also involves Relative formula mass

Potassium hydroxide reacts with hydrochloric acid to form potassium chloride and water.

potassium hydroxide + hydrochloric acid maths symbol potassium chloride + water

(a)

A student carried out a titration to find the exact volume of dilute hydrochloric acid that reacted with 25.0 cm3 of potassium hydroxide solution.

There were five steps in the titration. The steps shown are not in the correct order.

step J pour the potassium hydroxide solution into a conical flask and add a few drops of indicator to this solution step K fill a burette with the dilute hydrochloric acid and record the initial reading from the burette step L use a measuring cylinder to obtain 25 cm3 of potassium hydroxide solution step M take a final reading from the burette and calculate the volume of the dilute hydrochloric acid reacted step N run the dilute hydrochloric acid from the burette into the conical flask until the indicator changes colour

(a)(i)

Write the steps in the correct order.

Some of the steps have been completed for you.

first step
last step

Figure for question 4
[1]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

Suggest an alternative piece of apparatus that could be used in step L to obtain exactly 25.0 cm3 of potassium hydroxide solution.

[1]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(b)

A student was then asked to produce a pure sample of solid potassium chloride.

After finding the volume of acid reacted in step M, the student added this volume of acid to a fresh 25.0 cm3 sample of the potassium hydroxide solution. This mixture was then evaporated.

(b)(i)

Explain why this new mixture was evaporated rather than the original mixture from the titration, to produce a pure sample of solid potassium chloride.

[2]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

After evaporation, the mass of the potassium chloride was determined.

The theoretical yield of the experiment was 0.70 g. The actual yield was 0.84 g.

This gave a percentage yield greater than 100%.

Calculate the percentage yield of this experiment.

[2]
Reveal mark scheme Mark scheme for question 4(b)(ii)
(b)(iii)

Suggest a reason why the actual yield was greater than the theoretical yield.

[1]
Reveal mark scheme Mark scheme for question 4(b)(iii)
(b)(iv)

The equation for the reaction between potassium hydroxide solution and dilute hydrochloric acid is

Question 4, shown as an image

Calculate the atom economy for the production of potassium chloride from potassium hydroxide and hydrochloric acid. (relative formula masses: KOH = 56.0, HCl = 36.5, KCl = 74.5, H2O = 18.0)

Give your answer to one decimal place.

[4]
Reveal mark scheme Mark scheme for question 4(b)(iv)

Paper 1 June 2019 Higher tier, Question 4 11 marks

Calcium carbonate decomposes on heating to form calcium oxide and carbon dioxide.

Question 4, shown as an image
(a)

8.000 g of CaCO3 was heated strongly for about 10 minutes. 6.213 g of solid remained. Calculate the mass of carbon dioxide gas given off.

[1]
Reveal mark scheme Mark scheme for question 4(a)
(b)

A second sample of calcium carbonate is strongly heated in a crucible until there is no further loss in mass. The mass of calcium oxide remaining in the crucible is 5.450 g.

(b)(i)

The theoretical yield of calcium oxide in this experiment is 5.600 g.

Calculate the percentage yield of calcium oxide.

[2]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

The mass of solid left in the crucible is less than the theoretical mass of calcium oxide that should be obtained.

A possible reason for this is that

Question 4, shown as an image
[1]
Reveal mark scheme Mark scheme for question 4(b)(ii)
(c)

Another sample of calcium carbonate is heated and the mass of solid remaining is measured each minute.

The results are shown in Figure 3.

Figure for question 4
(c)(i)

Explain the trend shown by the data in Figure 3.

[2]
Reveal mark scheme Mark scheme for question 4(c)(i)
(c)(ii)

It is impossible to be sure from this data that the reaction is complete.

State why.

[1]
Reveal mark scheme Mark scheme for question 4(c)(ii)
(d)(i)

Calculate the relative formula mass of calcium carbonate, CaCO3. (relative atomic masses: C = 12, O = 16, Ca = 40)

[2]
Reveal mark scheme Mark scheme for question 4(d)(i)
(d)(ii)

Calculate the atom economy for the formation of calcium oxide in this reaction.

Question 4, shown as an image

You must show your working. (relative atomic masses: C = 12, O = 16, Ca = 40; relative formula mass: calcium oxide = 56)

[2]
Reveal mark scheme Mark scheme for question 4(d)(ii)

Paper 1 June 2018 Foundation tier, Question 10 11 marks

(a) Hydrogen burns in air at a temperature well above 100 Β°C to form water.

(a)
(a)(i)

The boiling points of hydrogen and water are shown in Figure 15.

Figure for question 10

Use this information to add the missing state symbols to the equation for the reaction taking place as the hydrogen burns.

Question 10, shown as an image
[2]
Reveal mark scheme Mark scheme for question 10(a)(i)
(a)(ii)

The atom economy for the reaction in (i) is 100%.

State how the equation shows that the atom economy is 100%.

[1]
Reveal mark scheme Mark scheme for question 10(a)(ii)
(b)

Lead can be obtained by heating its oxide with carbon. The balanced equation for the reaction is

Question 10, shown as an image

Calculate the atom economy for the production of lead in this reaction. (relative atomic masses: C = 12, O = 16, Pb = 207 relative formula masses: PbO = 223, CO2 = 44)

Give your answer to two significant figures.

[4]
Reveal mark scheme Mark scheme for question 10(b)
(c)(i)

In an experiment to produce lead, 7.67 g of lead are obtained. The theoretical yield of lead for the experiment is 11.80 g.

Calculate the percentage yield of lead in this experiment.

[2]
Reveal mark scheme Mark scheme for question 10(c)(i)
(c)(ii)

In most reactions, the percentage yield of any product is less than 100%.

Give two reasons why the percentage yield is less than 100%.

[2]
Reveal mark scheme Mark scheme for question 10(c)(ii)

Paper 1 June 2018 Higher tier, Question 4 11 marks

Also involves The three states of matter

(a) Hydrogen burns in air at a temperature well above 100 maths symbolC to form water.

(a)
(a)(i)

The boiling points of hydrogen and water are shown in Figure 3.

boiling point in C
hydrogenβˆ’253
water100
Question 4, shown as an image

Use this information to add the missing state symbols to the equation for the reaction taking place as the hydrogen burns.

Question 4, shown as an image
[2]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

The atom economy for the reaction in (i) is 100%.

State how the equation shows that the atom economy is 100%.

[1]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(b)

Lead can be obtained by heating its oxide with carbon. The balanced equation for the reaction is

Question 4, shown as an image

Calculate the atom economy for the production of lead in this reaction. (relative atomic masses: C = 12, O = 16, Pb = 207 relative formula masses: PbO = 223, CO2 = 44)

Give your answer to two significant figures.

[4]
Reveal mark scheme Mark scheme for question 4(b)
(c)(i)

In an experiment to produce lead, 7.67 g of lead are obtained. The theoretical yield of lead for the experiment is 11.80 g.

Calculate the percentage yield of lead in this experiment.

[2]
Reveal mark scheme Mark scheme for question 4(c)(i)
(c)(ii)

In most reactions, the percentage yield of any product is less than 100%.

Give two reasons why the percentage yield is less than 100%.

[2]
Reveal mark scheme Mark scheme for question 4(c)(ii)

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