Question

Mg was reacted with an excess of $HCl$ dilute and the $H_2$ the gas produced collected in a eudiometer. The volume of hydrogen in the eudiometer was corrected to conditions of STP. If 94.1 millilitres of $H_2$ was produced, how much Mg reacted in this reaction?

• A. 0.2 gm
• B. 0.1 gm
• C. 0.3 gm
• D. 0.4 gm

Answer 1

The correct option is B 0.1 gm

Volume of $H_2$ produced $=94.1ml$

Moles of $H_2$ produced $=\frac{V}{V_m}=\frac{94.1}{22400}=4.2\times {10}^{-3}mol$

Now,

$Mg+2HCl\to MgC{l}_2+H_2$

1 mol of Mg is consumed when 1 mol of $H_2$ is produced.

So, $4.2\times {10}^{-3}$ mol of Mg is consumed when $4.2\times {10}^{-3}$ mol of $H_2$ is produced.

Mass of Mg consumed $=n\times M=4.2\times {10}^{-3}\times 24=0.1g.$