The volume of hydrogen gas produced at S.T.P. when \(100 ml\) of \(2 M\) hydrochloric acid reacts with exeess zine is ____________
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Correct Answer: Option A
Explanation:
Equation of reaction
\begin{aligned}
&Zn _{(b)}+2 HCl _{(2 q)} \rightarrow ZnCl _{2(\text { aq) }}+ H _{2( g )}\\
&\eta HCl =\text { conc. } \times \text { vol }\\
&100 ml =100 cm ^{3}=0.1 dm ^{3}\\
&\eta HCl =2 \times 0.1=0.2 mol\\
&\text { from the equation, }\\
&2 \text { mole } HCl \rightarrow 1 \text { mole } H _{2}\\
&0.2 mol HCl \rightarrow x\\
&x=\frac{0.2}{2}=0.1 mol H _{2}\\
&\text { at S.T.P., }\\
&1 mole -22.4 dm ^{3}\\
&0.1 \text { mole }-y\\
&y=2.24 dm ^{3} \text { of } H _{2}
\end{aligned}
Equation of reaction
\begin{aligned}
&Zn _{(b)}+2 HCl _{(2 q)} \rightarrow ZnCl _{2(\text { aq) }}+ H _{2( g )}\\
&\eta HCl =\text { conc. } \times \text { vol }\\
&100 ml =100 cm ^{3}=0.1 dm ^{3}\\
&\eta HCl =2 \times 0.1=0.2 mol\\
&\text { from the equation, }\\
&2 \text { mole } HCl \rightarrow 1 \text { mole } H _{2}\\
&0.2 mol HCl \rightarrow x\\
&x=\frac{0.2}{2}=0.1 mol H _{2}\\
&\text { at S.T.P., }\\
&1 mole -22.4 dm ^{3}\\
&0.1 \text { mole }-y\\
&y=2.24 dm ^{3} \text { of } H _{2}
\end{aligned}