Key Concept: Electrolysis of Water, Gas Formation
c) 6.72 L
[Solution Description]
First, calculate the moles of water decomposed:
$\text{Moles of } H_2O = \frac{3.6 \, \text{g}}{18 \, \text{g/mol}} = 0.2 \, \text{mol}$
From the balanced equation $2H_2O \rightarrow 2H_2 + O_2$, 2 moles of $H_2O$ produce 2 moles of $H_2$ and 1 mole of $O_2$. Therefore,
$\text{Moles of } H_2 = 0.2 \, \text{mol}, \quad \text{Moles of } O_2 = 0.1 \, \text{mol}$
Total moles of gas produced:
$\text{Total moles} = 0.2 + 0.1 = 0.3 \, \text{mol}$
Volume of gases at STP:
$\text{Volume} = 0.3 \, \text{mol} \times 22.4 \, \text{L/mol} = 6.72 \, \text{L}$
Your Answer is correct.
c) 6.72 L
[Solution Description]
First, calculate the moles of water decomposed:
$\text{Moles of } H_2O = \frac{3.6 \, \text{g}}{18 \, \text{g/mol}} = 0.2 \, \text{mol}$
From the balanced equation $2H_2O \rightarrow 2H_2 + O_2$, 2 moles of $H_2O$ produce 2 moles of $H_2$ and 1 mole of $O_2$. Therefore,
$\text{Moles of } H_2 = 0.2 \, \text{mol}, \quad \text{Moles of } O_2 = 0.1 \, \text{mol}$
Total moles of gas produced:
$\text{Total moles} = 0.2 + 0.1 = 0.3 \, \text{mol}$
Volume of gases at STP:
$\text{Volume} = 0.3 \, \text{mol} \times 22.4 \, \text{L/mol} = 6.72 \, \text{L}$