What volume of oxygen will be released during electrolysis of 750 g of water containing 3% impurities?

1. Let’s compose the equation of water electrolysis:

2H2O → 2H2 ↑ + O2 ↑;

K (-): 2H + + 2 e- → H20;

A (+): 2O- – 2 e- → O2;

2.Calculate the mass of water free of impurities:

m (H2O) = m (portions) – m (impurities);

m (impurities) = m (portions) * w (impurities) = 750 * 0.03 = 22.5 g;

m (H2O) = 750-22.5 = 727.5 g;

3. find the chemical amount of water:

n (H2O) = m (H2O): M (H2O) = 727.5: 18 = 40.4 mol;

4.Set the amount of released oxygen:

n (O2) = n (H2O): 2 = 40.4: 2 = 20.2 mol;

5.calculate the volume of oxygen:

V (O2) = n (O2) * Vm = 20.2 * 22.4 = 452.48 liters.

Answer: 452.48 liters.