Fig. 1 shows a simplified diagram of an electrolytic cell used by a recycling company. Molten sodium chloride is placed in the heated container. The company...
Question 1 Report
Fig. 1 shows a simplified diagram of an electrolytic cell used by a recycling company. Molten sodium chloride is placed in the heated container. The company wants to calculate the mass of sodium metal that could be obtained from a pure 11.7 g sample of sodium chloride. The relative atomic masses are Na = 23 and Cl = 35.5.
(a) Name the metal made at the negative electrode. [1]
(b) Explain why sodium chloride must be molten rather than solid. [1]
(c) Use the formula NaCl to calculate its relative formula mass. [1]
(d) Use 11.7 g of NaCl to calculate the mass of sodium that can be made. [2]
The table below shows data from a water-treatment company. The company neutralises a pure sample of hydrochloric acid using sodium hydroxide. The equation is HCl + NaOH → NaCl + H2O. The relative formula mass of sodium hydroxide is 40.0. A fixed volume of acid is used each time, and the sodium hydroxide is weighed as a dry solid before making a solution.
| trial | mass of NaOH used / g | amount of NaOH / mol |
|---|---|---|
| 1 | 1.00 | 0.0250 |
| 2 | 2.00 | 0.0500 |
| 3 | 3.00 | 0.0750 |
In trial 2, exactly 0.0500 mol of acid reacts.
(a) Name the salt made in this reaction. [1]
(b) Use the equation to state the amount of HCl reacting in trial 2. [1]
(c) Complete this calculation for trial 3: amount = mass ÷ Mr. [1]
(d) Explain why 2.00 g of NaOH is needed when 0.0500 mol of HCl reacts. [2]
Answer Details
Electrolysis of molten sodium chloride
(a) The metal made at the negative electrode is sodium. Positive sodium ions gain electrons at the negative electrode. [1]
(b) Sodium chloride must be molten because its ions can move when molten. Moving ions carry charge, so the molten liquid conducts electricity. In solid sodium chloride, the ions are fixed in a lattice. [1]
(c) \[M_r(\mathrm{NaCl})=23+35.5=58.5\]
The relative formula mass is 58.5. [1]
(d) Calculate the amount of sodium chloride:
\[\text{amount of NaCl}=\frac{11.7\text{ g}}{58.5\text{ g mol}^{-1}}=0.200\text{ mol}\]
Each formula unit of NaCl contains one sodium ion, so 0.200 mol of NaCl can produce 0.200 mol of sodium.
\[\text{mass of Na}=0.200\text{ mol}\times23\text{ g mol}^{-1}=4.60\text{ g}\]
The mass of sodium is 4.60 g. [2]
Neutralisation with sodium hydroxide
(a) The salt made is sodium chloride. [1]
(b) The equation \(\mathrm{HCl+NaOH\rightarrow NaCl+H_2O}\) has a 1:1 mole ratio between HCl and NaOH. Thus, in trial 2, the amount of HCl reacting is 0.0500 mol. [1]
(c) \[\text{amount of NaOH}=\frac{3.00\text{ g}}{40.0\text{ g mol}^{-1}}=0.0750\text{ mol}\]
The calculation for trial 3 gives 0.0750 mol. [1]
(d) The equation shows a 1:1 mole ratio, so 0.0500 mol of HCl requires 0.0500 mol of NaOH for complete neutralisation. The corresponding mass is:
\[\text{mass of NaOH}=0.0500\text{ mol}\times40.0\text{ g mol}^{-1}=2.00\text{ g}\]
Therefore, 2.00 g of NaOH is needed. [2]
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