The drain cleaner contains small bits of aluminium which reacts with caustic soda to produce hydrogen gas. What volume of $H_{2}$ gas at $20^{0} C$ and $1$ bar pressure will be released when $0.27 g$ aluminium reset with caustic soda. $(a t . m a s s o f A l = 27 u)$ $2 A l + 2 N a O H + 2 H_{2} O \to 2 N a A l O_{2} + 3 H_{2}$

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Solution

$2 A l + 2 N s O H + 2 H_{2} O \to 2 N a A l O_{2} + 3 H_{2}$
number of Aluminium $= \frac{0.27}{27} = 0.01 m o l e$
$2$ mole aluminium react with caustic soda and give $3 m o l H_{2}$
$0.001$ mole aluminium react with caustic soda and give $\frac{3}{2} \times 0.01$
$n = \frac{0.08}{2} = 0.015$
Temperature of $H_{2} T = 20^{o} + 273 = 293 k$
present $P = 1 b a r, R = 8.31 \times 10^{- 2} 2 b a r k^{- 1} m o l e^{- 1}$
$V = \frac{n R T}{P} l i t r e$
$= \frac{0.015 \times 8.314 \times 10^{- 2} \times 293}{1 h o u r}$ litre
$= 0.3659$ litre $= 365.9 m l$

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