Mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dil. HCl is…
Given : Molar mass of Mg is $24 \mathrm{~g} \mathrm{~mol}^{-1}$.
- 235.7 g
- 0.24 mg
- 236 mg
- 2.444 g
Solution
Volume $\mathrm{H}_2$ evolved $=220 \mathrm{ml}$
Mole of $\mathrm{H}_2=\frac{220 \times 10^{-3}}{22.4}=$ mole of Mg used
$\begin{aligned}
\therefore \text { Mass of Mg used } & =\frac{220 \times 10^{-3}}{22.4} \times 24 \\
& =235.7 \times 10^{-3} \mathrm{gm} \\
& =235.7 \mathrm{mg}
\end{aligned}$
Asked in: JEE Main 2025 (03 Apr Shift 2)
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