Question

What volume of hydrogen, measured at STP was liberated when $1.16\mathrm{g}$ of magnesium was allowed to react with excess dilute sulfuric acid?

Answer 1

Step 1: Given information

• Mass of Magnesium $\left(\mathrm{Mg}\right)=1.16\mathrm{g}$

Step 2: Reaction involved

• Magnesium $\left(\mathrm{Mg}\right)$ reacts with dilute sulfuric acid $\left({\mathrm{H}}_2{\mathrm{SO}}_4\right)$ to form magnesium sulfate $\left({\mathrm{MgSO}}_4\right)$ and hydrogen $\left({\mathrm{H}}_2\right)$ gas.
• The reaction is as follows:

$\mathrm{Mg}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{MgSO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

Step 3: Volume of gas at STP

• The balanced chemical reaction is as follows:

$\mathrm{Mg}\left(\mathrm{s}\right)+{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)\to {\mathrm{MgSO}}_4\left(\mathrm{aq}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)$

$\begin{array}{rcl}\mathrm{Mass}\mathrm{of}\mathrm{Mg}& =& 24\mathrm{g}\end{array}$

Volume of 1 mole of a gas is $22.4{\mathrm{dm}}^3$ at STP.

Step 4: Calculate the volume of Hydrogen

• $24\mathrm{g}$ of magnesium liberates $22.4{\mathrm{dm}}^3$ of hydrogen gas at STP (Standard temperature and pressure).
• Thus, $1.16\mathrm{g}$ of magnesium will liberate $\frac{22.4{\mathrm{dm}}^3}{24\mathrm{g}}\times 1.16\mathrm{g}=1.083{\mathrm{dm}}^3$ of hydrogen gas at STP.

Therefore, $1.083{\mathrm{dm}}^3$ of hydrogen was liberated when $1.16\mathrm{g}$ of magnesium was allowed to react with excess dilute sulfuric acid at STP.