The table below shows data from a ceramics company testing calcium carbonate in a glaze. The carbonate reacts with nitric acid before the glaze is fired. Fi...
Question 1 Report
The table below shows data from a ceramics company testing calcium carbonate in a glaze. The carbonate reacts with nitric acid before the glaze is fired. Fig. 1 shows the company using a pipette to add 25.0 cm3 portions of acid to a flask containing powdered carbonate.
| measurement | value |
|---|---|
| concentration of HNO3 | 0.400 mol dm-3 |
| volume of HNO3 | 25.0 cm3 |
| mass of CaCO3 sample | 1.50 g |
The equation is CaCO3 + 2HNO3 → Ca(NO3)2 + H2O + CO2. Mr of CaCO3 is 100.
(a) Calculate the amount, in mol, of nitric acid added. [1]
(b) Use the equation to calculate the maximum amount, in mol, of calcium carbonate that can react. [1]
(c) Calculate the mass of calcium carbonate left unreacted. [2]
(d) Give the name of the reactant that is used up completely. [1]
Fig. 1 shows an aluminium drinks-can recycling trial. Pieces of cleaned aluminium are added to aqueous sodium hydroxide, and hydrogen is collected over water. The reaction is 2Al + 2NaOH + 6H2O → 2NaAl(OH)4 + 3H2. The table below gives results from separate runs. Relative atomic mass of aluminium is 27.0.
| run | mass of Al / g | volume H2 / cm3 |
|---|---|---|
| 1 | 0.540 | 720 |
| 2 | 0.270 | 358 |
(a) Calculate the amount, in mol, of aluminium in run 1. [1]
(b) Use the equation to calculate the expected volume of hydrogen for run 1, at room conditions. [3]
(c) Give one reason why the measured volume in run 2 is slightly less than the calculated value. [1]
Answer Details
Calcium carbonate and nitric acid
- \[25.0\text{ cm}^3=0.0250\text{ dm}^3\]
\[n(\mathrm{HNO_3})=0.400\times0.0250=0.0100\text{ mol}\]
[1] - Two moles of nitric acid react with one mole of calcium carbonate:
\[n(\mathrm{CaCO_3})=\frac{0.0100}{2}=0.00500\text{ mol}\]
[1] - \[n(\mathrm{CaCO_3\ initially})=\frac{1.50}{100}=0.0150\text{ mol}\]
\[m(\mathrm{unreacted\ CaCO_3})=(0.0150-0.00500)\times100=1.00\text{ g}\]
[2] - Nitric acid is used up completely, so it is the limiting reactant. [1]
Aluminium recycling
- \[n(\mathrm{Al})=\frac{0.540}{27.0}=0.0200\text{ mol}\]
[1] - The equation gives \(2\mathrm{Al}:3\mathrm{H_2}\):
\[n(\mathrm{H_2})=0.0200\times\frac{3}{2}=0.0300\text{ mol}\]
\[V=0.0300\times24=0.720\text{ dm}^3=720\text{ cm}^3\]
[3] - The lower measured volume may be due to gas loss from the apparatus, incomplete reaction, slight hydrogen dissolution, or a reading error. [1]
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