What volume of oxygen will be released during electrolysis of 750 g of water containing 3% impurities?
August 29, 2021 | education|
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.
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