Oxford Revise AQA GCSE Chemistry | Chapter C7 answers

C7: Chemical calculations with moles

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Answers

Extra information

Mark

AO / Specification reference

01.1

percentage yield:

\(
\frac{{{\rm{mass\ of\ product\ actually\ made}}}}{{{\rm{maximum\ theoretical\ mass\ of\ product}}}} \times 100\%
\)

atom economy:

\(
\frac{{{\rm{relative\ formula\ mass\ of\ desired\ product\ from\ equation}}}}{{{\rm{sum\ of\ relative\ formula\ masses\ of\ all\ reactants\ from\ equations}}}} \times 100\%
\)

1

1

AO1

4.3.3.1

4.3.3.2

01.2

two from:

  • sustainable development/preserves Earth’s resources
  • economic reasons
  • reduce waste

1

1

AO1

4.3.3.2

01.3

some remains on the filter paper/is not scraped off

1

AO3

02.1

two from:

  • wear eye protection
  • use a safety screen between students and reaction
  • stand back immediately when reaction starts

allow any other suitable precaution

1

1

AO3

02.2

\(
\begin{array}{l}
{{\rm{M}}_{\rm{r}}}{\rm{of\ iron}}\left( {{\rm{III}}} \right){\rm{\ oxide\ is\ }}\left( {2 \times 56} \right) + \left( {3 \times 16} \right) = 160{\rm{\ g}}\\
8.0{\rm{\ g\ of\ iron}}\left( {{\rm{III}}} \right){\rm{\ oxide\ is\ }}\frac{8}{{160}} = 0.050{\rm{\ mol}}\\
2.7{\rm{\ g\ of\ aluminium\ is\ }}\frac{{2.7}}{{27}} = 0.10{\rm{\ mol}}
\end{array}
\)

from balanced equation, one mol of iron(III) oxide reacts with two mol of aluminium, so 0.050 mol of iron(III) oxide needs 0.10 mol of aluminium.

1

1

1

1

AO2

4.3.2.1

4.3.2.4

02.3

from balanced equation, one mol of iron(III) oxide makes two mol of iron,

so 0.050 mol of iron(III) oxide makes 0.10 mol of iron

this has a mass of 0.10 × 56 = 5.6 g

1

1

AO2

4.3.2.1

02.4

percentage yield:

\(
\begin{array}{l}
\frac{{{\rm{mass\ of\ product\ actually\ made}}}}{{{\rm{maximum\ theoretical\ mass\ of\ product}}}} \times 100\% \\
\frac{{4.6}}{{5.6}} \times 100 = 82.1\%
\end{array}
\)

allow error carried forward

1

AO1 × 1

AO2 × 1

4.3.3.1

02.5

some of the aluminium reacts with oxygen from the air

some of the iron made is not collected

allow other suitable reasons

1

1

AO3

4.3.3.1

03.1

\(
\begin{array}{l}
{\rm{number\ of\ moles\ of\ NaOH\ }} = \frac{{25}}{{1000}} \times 0.100 = 0.00250\\
{\rm{from\ balanced\ equation,\ one\ mol\ of\ }}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}{\rm{\ reacts\ with\ two\ mol\ of\ NaOH,}}\\
{\rm{so\ number\ of\ moles\ of\ acid\ in\ 25}}{\rm{.0\ c}}{{\rm{m}}^{\rm{3}}} = \frac{{0.00250}}{2} = 0.00125{\rm{\ mol}}\\
{\rm{concentration\ of\ acid\ }} = 0.00125 \times \frac{{1000}}{{25}} = 0.05{\rm{\ mol/d}}{{\rm{m}}^3}\\
= 0.0500{\rm{\ mol\ /d}}{{\rm{m}}^{\rm{3}}}
\end{array}
\)

1

1

1

1

1

AO1 × 1

AO2 × 4

4.3.4

03.2

Mr of H2SO4 = (2 × 1) + 32 + (4 × 16) = 98 g

mass of 0.0500 mol = 0.0500 × 98 g = 4.9 g,

so concentration = 4.9 g/dm3

1

1

AO2

4.3.2.1

03.3

Mr of NaOH = 23 + 16 + 1 = 40 g

mass of 0.0100 mol = 40 × 0.100 = 4.0 g

1

1

AO2

4.3.2.1

04.1

Mr of C2H5OH = (2 × 12) + (5 × 1) + 16 + 1 = 46

1

AO2

4.3.1.2

04.2

\(
\begin{array}{l}
{\rm{atom\ economy\ of\ process\ }}1 = \frac{{46}}{{(28 + 18)}} \times 100 = 100\% \\
{\rm{atom\ economy\ of\ process\ }}2 = \;\frac{{(2 \times 46)}}{{180}} \times 100 = 51.1\%
\end{array}
\)

the atom economy process of 1 is approximately double that of process 2

1

1

1

AO2

4.3.3.2

04.3

Level 3: The comparisons are detailed and accurate. The writing is clear, coherent and logical and comparisons are clearly made.

5-6

AO3

4.3.3.2

Level 2: The comparisons are generally correct, although may lack detail. The writing is mainly clear, although the structure may lack logic and comparisons are not always clear.

3-4

Level 1: Some comparisons are correct. The writing lacks clarity, coherence and logic, and the comparisons are not clearly expressed.

1-2

No relevant content

0

Indicative content

  • 1 occurs at a higher temperature and pressure than 2, so 2 is better for sustainable development in this respect
  • the raw material for 1 is obtained from crude oil, so 2 is better for sustainable development in this respect
  • 2 produces carbon dioxide, which is a greenhouse gas, so 1 is better for sustainable development in this respect
  • 1 has a higher atom economy than 2, so 1 is better for sustainable development in this respect

05.1

Mr = (3 × 12) + (8 × 1) = 44

\(
{\rm{number\ of\ moles\ }} = \frac{{6000}}{{44}} = 136{\rm{\ mol}}
\)

at room temperature and pressure, one mol of gas occupies 24 dm3

136 mol occupies 136 × 24 = 3264 dm3

1

1

1

1

AO1 × 1

AO2 × 3

4.3.5

05.2

50 × 5 = 250 cm3

\(
\frac{{250}}{{1000}} = 0.250{\rm{\ d}}{{\rm{m}}^3}
\)

1

1

AO1 × 1

AO2 × 1

4.3.5

05.3

\(
{\rm{number\ of\ moles\ of\ propane\ }} = \frac{{480}}{{44}} = 10.9
\)

from balanced equation, one mol of propane makes three mol of CO2

number of mol of CO2 = 3 × 10.9 = 32.7 mol

24 × 32.7 = 784.8 dm3

= 785 dm3

1

1

1

1

AO1 × 2

AO2 × 3

4.3.5

06.1

8.8 g

1

AO2

4.3.1.1

4.3.3.1

06.2

8.2 g

1

AO2

4.3.3.1

06.3

100%

1

AO2

4.3.3.2

07

Level 3: Appropriate equipment named and a detailed description of the various repeats required is provided.

5-6

AO1

4.4.2.5

Level 2: Method provided. Some attempt at demonstrating need for repeats.

3-4

Level 1: A basic titration method provided. No mention of repeats.

1-2

No relevant content.

0

Indicative content

  • use a pipette to measure out a known volume of sodium hydroxide.
  • put the sodium hydroxide into a conical flask.
  • add a few drops of a suitable indicator to the conical flask
  • place the conical flask on a white tile.
  • fill a burette with the hydrochloric acid.
  • add about one m3 of acid to the conical flask and mix by swirling the flask.
  • repeat until the indicator changes colour.
  • record the volume of acid used as the rough titre.
  • repeat the process, but as the end point is approached, add the acid drop wise to obtain a precise measurement.
  • repeat until at least two concordant results are achieved.

08.1

Mr of CH4 = 12 + (4 × 1) = 16

Mr of H2O = (1 × 2) + 16 = 18

atom economy:

\(
\begin{array}{l}
\frac{{{\rm{relative\ formula\ mass\ of\ desired\ product\ from\ equation}}}}{{{\rm{sum\ of\ relative\ formula\ masses\ of\ all\ reactants\ from\ equations}}}} \times 100\% \\
\frac{6}{{(16 + 18)}} \times 100\\
= 17.6\%
\end{array}
\)

1

1

1

AO1 × 1

AO2 × 2

4.3.3.2

08.2

use electricity generated from renewable resources

allow suitable alternative answers

1

AO3

08.3

Level 3: The comparisons are detailed and accurate. The writing is clear, coherent and logical and comparisons are clearly made.

5-6

AO3

4.3.3.2

Level 2: The comparisons are generally correct, although may lack detail. The writing is mainly clear, although the structure may lack logic and comparisons are not always clear.

3-4

Level 1: Some comparisons are correct. The writing lacks clarity, coherence and logic, and the comparisons are not clearly expressed.

1-2

No relevant content

0

Indicative content

  • 1 occurs at a higher temperature than 2, so 2 is better for sustainable development in this respect
  • if the raw material for 1 is obtained from fossil fuels, 2 is better for sustainable development in terms of resources used
  • if the material for 1 is obtained from sewage, both processes have a similar impact on the environment in terms of resources used
  • 2 produces carbon monoxide, which is poisonous, so 1 is better for sustainable development in terms of pollutants made
  • 1 has a higher atom economy than 2, so 1 is better for sustainable development in this respect

09.1

to allow oxygen to enter the crucible

1

AO3

09.2

percentage yield:

\(
\begin{array}{l}
\frac{{{\rm{mass\ of\ product\ actually\ made}}}}{{{\rm{maximum\ theoretical\ mass\ of\ product}}}}\ \times\ 100\% \\
\frac{1.80}{2.00}\ \times\ 100\% \\
= 90\%
\end{array}
\)

1

1

1

AO1 × 1

AO2 × 2

4.3.3.1

09.3

one from:

  • some magnesium oxide escaped out of the crucible
  • not all the magnesium reacted
  • some of the magnesium oxide reacted with nitrogen from the air

allow other suitable answers

1

AO3

4.3.3.1

10.1

13.55

1

AO3

4.4.2.5

10.2

13.00

1

AO2

4.4.2.5

10.3

HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)

1 mark for reactants, 1 mark for products, 1 mark for state symbols

3

AO2

4.4.2.5

10.4

converting units, 25 cm3 = 0.025 dm3 and 13 cm3 = 0.013 dm3

moles of NaOH = 0.1 × 0.025 = 2.5 × 10–3

\(
\begin{array}{l}
{\rm{concentration\ of\ HN}}{{\rm{O}}_3} = \frac{{2.5 \times {{10}^{-8}}}}{{0.013}}\\
= 0.19{\rm{\ mol/d}}{{\rm{m}}^3}
\end{array}
\)

accept errors carried forward for ratios from question 10.3

1

1

1

1

AO1

AO2

4.4.2.5

11.1

F

1

AO1

4.1.2.1

4.1.2.5

11.2

two from

  • A
  • B
  • C
  • D

two correct letters required for the mark

1

AO1

4.1.2.3

11.3

Level 3: The description is detailed and accurate. The writing is clear, coherent and logical.

5-6

AO1

4.1.2.2

Level 2: The description is correct, although lacks detail. The writing is mainly clear, although the structure may lack logic.

3-4

Level 1: Some aspects of the description are correct. The writing lacks clarity, coherence and logic.

1-2

No relevant content

0

Indicative content

  • before discovering sub-atomic particles, scientists attempted to classify the elements by arranging them in order of their atomic weights
  • early periodic tables were incomplete, and some elements were placed in inappropriate groups
  • Mendeleev overcame the problems by leaving gaps for elements that he thought had not been discovered
  • Mendeleev also changed the order of elements in some places based on atomic weights
  • elements predicted by Mendeleev were discovered and filled the gaps
  • knowledge of isotopes made it possible to explain why the order based on atomic weights was not always correct

12.1

A

1

AO3

4.2.1.4

12.2

one dot and one cross in each of the four intersections

2

AO1

4.2.1.4

12.3

B

ionic bonding, no free electrons

only able to conduct electricity when molten because ions can move

1

1

1

AO3

12.4

metallic

giant structure of atoms/ions arranged in regular pattern

electrons in the outer shell of metal atoms are delocalised and free to move throughout structure

giving rise to strong metallic bonds

1

1

1

1

AO1

4.2.1.5

13.1

55.6 × 5 = 278 mol

1

AO2

4.3.2.5

13.2

Mr = (6 × 12) + (12 × 1) + (6 × 16) = 180 g

concentration = 180 × 300

= 59 400 g/dm3

= 59.4 kg/dm3

1

1

1

1

AO1 × 2

AO2 × 2

4.3.2.1

4.3.2.5

13.3

\(
\begin{array}{l}
59.4\ \times\ \frac{{50}}{{1000}}\\
= 2.97\rm{\ kg}
\end{array}
\)

award two marks for correct answer without working

allow 2970 g

1

1

AO2

4.3.2.5

14.1

heat the solution

until the water evaporates

leaving potassium chloride crystals

1

1

1

AO1

4.1.1.2

14.2

whole square filled with same-sized circle

circles arranged in regular pattern

all circles touching

1

1

1

AO1

4.2.2.1

14.3

potassium chloride has different properties as a compound to potassium and chlorine elements

allow named properties of K and Cl e.g., colour, electrical conductivities

1

AO1

4.1.1.2

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