$1$ litre of oxygen and $1$ litre of hydrogen are taken in a vessel of $2$ -litre capacity at $N T P$ . The gases are made to combine by applying electric sparks. Assume that water is formed quantitatively. How many grams of water are formed? What is the other component present in the vessel and in what weight? If the vessel is now heated to $100^{0} C$ , what will be the pressure inside the vessel in mmHg?

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Solution

$V (H_{2})$ $= 1 l i t$ & $V (O_{2})$ $= 1 l i t$

Before reaction,
$2 H_{2} 1 l i t + O_{2} 1 l i t \to 2 H_{2} O (l) 0$

After reaction,
$2 H_{2} 0 + O_{2} 1 - 1 / 2 l i t \to 2 H_{2} O (l) 1 l i t$

$∴$ Volume of $O_{2}$ remained = $\frac{1}{2}$ lit

$∴$ Mole of $O_{2} = \frac{\frac{1}{2}}{22.4} = \frac{1}{44.8}$

$W t . o f O_{2} = \frac{1}{48.8} \times 32 g = 0.7143 g$

Also moles of $H_{2} O$ produced from $\frac{1}{22.4}$ moles of $H_{2} = \frac{1}{22.4}$

$∴ W t p r o d u c e d = \frac{1}{22.4} \times 18 = 0.8036 g$

At $100^{0} C$ , water becomes gas

$∴$ Total volume of gases = $1 + \frac{1}{2} = \frac{3}{2} l i t$

$∴ P V = n R T$

$P = \frac{n R T}{V} = \frac{(\frac{1}{44.8} + \frac{1}{22.4}) \times 0.0821 \times 373}{\frac{3}{2}}$

$= 1.34 a t m = 1018.4 m m o f H g$

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