Using the molar volume in calculations with chemical equations
3 marks · EasyThe table shows the volume of a gas produced from the reaction of different masses of magnesium with excess hydrochloric acid at room temperature and pressure.
| Mass of magnesium (g) | Volume of H₂ gas (dm³) |
|---|---|
| 0.12 | 0.12 |
| 0.24 | 0.24 |
| 0.36 | 0.36 |
Describe the relationship between the mass of magnesium and the volume of hydrogen gas produced.
Suggest the volume of hydrogen gas (in dm³) produced from 0.48 g of magnesium.
Explain your answer to part (b).
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Directly proportional · must show States direct proportionality or linear relationship
(b) 0.48 dm³
(c) Because for every gram of magnesium used, the same volume in dm³ of hydrogen is produced. · must show Explains by reference to pattern in table (mass = volume)
Common mistakes: Not recognising proportionality; Guessing part (b)
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Marking points
B1 (a) · 1 mark · Recognises and states that mass of Mg and volume of H₂ are directly proportional
A1 (b) · 1 mark · Correctly predicts 0.48 dm³ for 0.48 g of Mg
B2 (c) · 1 mark · Explains that the answer is because the volume (in dm³) always equals the mass (in g) in this data.
Worked solution
(a) The mass of magnesium and volume of hydrogen are directly proportional.
(b) The volume produced from 0.48 g of magnesium would be 0.48 dm³.
(c) This is because in each case, the number (in g) is the same as the volume (in dm³), so doubling the mass doubles the volume.