Problem 1

An amount of 20 g of potassium sulphate was dissolved in 150 cm

^3

of water. The solution was then electrolysed. After electrolysis, the content of potassium sulphate in the solution was 15 % by mass. What volumes of hydrogen and oxygen were obtained at a temperature of 20 °C and a pressure of 101 325 Pa?

Step 3 of 3: Convert to volumes

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V(\ce{H2}) = \dfrac{nRT}{p} = \dfrac{2.04 \times 8.314 \times 293.15}{101325} \approx 0.049\ \text{m}^3 \approx 49\ \text{dm}^3; \quad V(\ce{O2}) \approx 24.5\ \text{dm}^3

Analysis

The ideal gas law at 293.15 K and 101 325 Pa gives $V(\ce{H2}) \approx 49$ dm $^3$ and $V(\ce{O2}) = \tfrac{1}{2}V(\ce{H2}) \approx 24.5$ dm $^3$ .

Common pitfall. The sulphate is a spectator: do not try to balance an electrolysis of K2SO4 itself.