Q. The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen in mL at $20^\circ C$ and one bar will be released when 0.15 g of aluminium reacts? (Take $\text{1} \, \text{bar} \, \text{=} \, \text{0} \text{.987} \, \text{atm}$ )
NTA AbhyasNTA Abhyas 2020States of Matter
Solution:
The reaction between aluminium and caustic soda is
$\underset{\underset{= 54 g}{2 \times 27}}{2 A l}+2NaOH+2H_{2}O \rightarrow 2NaAlO_{2}+\underset{\underset{\text{atSTP}}{3 \times 22.4 L}}{3 H_{2}}$
So 54 g of Al produces $H_{2}$ at STP $=3\times 22.4$ L
0.15 g of Al will produce $H_{2}$ at STP
$=\frac{3 \times 22.4}{54}\times 0.15$
$=0.186L$
At STP Given Conditions $P_{1}=1$ atm $P_{2}=1$ bar $=0.987$ atm $V_{1}=0.186$ L $V_{2}=?$ $T_{1}=273$ K $T_{2}=273+20=293K$
Applying ideal gas equation, $\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$
$\frac{1 \times 0.1867}{273}=\frac{0.987 \times V_{2}}{293}$
$V_{2}=\frac{293}{0.987}\times \frac{1 \times 0.1867}{273}$
$=0.2030$ L $=203$ mL
| At STP | Given Conditions |
|---|---|
| $P_{1}=1$ atm | $P_{2}=1$ bar $=0.987$ atm |
| $V_{1}=0.186$ L | $V_{2}=?$ |
| $T_{1}=273$ K | $T_{2}=273+20=293K$ |