The drain cleaner, Drainex contains small bits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at $20^{\circ} C$ and one bar will be released when 0.15g of aluminum reacts?

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Solution

The reaction of aluminium with caustic soda can be represented as:
$2 A l + 2 N a O H + 2 H_{2} O \to 2 N a A l O_{2} + 3 H_{2} 2 \times 27 g 3 \times 22400 m L$
At STP (273.15 K and 1 atm), 54 g (2 $\times$ 27 g) of Al gives 3 $\times$ 22400 mL of $H_{2}$ ..
$∴ 0.15 g A l g i v e s \frac{3 \times 22400 \times 0.15}{54} m L o f H_{2} i . e ., 186.67 m L o f H_{2} .$
At STP,
$p_{1} = 1 a t m v_{1} = 186.67 m L T_{1} = 273.15 K$
Let the volume of dihydrogen be $V_{2} a t p_{2} = 0.987 a t m$ (since 1 bar = 0.987 atm) and $T_{2} = 20^{\circ} C = (273.15 + 20) K = 293.15 K$ .
Now,
$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{1} 2}{T_{2}} \Rightarrow V_{2} \frac{p_{1} V_{1} T_{2}}{p_{2} T_{1}} = \frac{1 \times 186.67 \times 293.15}{0.987 \times 273.15} = 202.98 m L = 203 m L$

Therefore, 203 mL of dihydrogen will be released.

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