Question

In the following reaction, what mass of magnesium sulfate is formed when $6\mathrm{g}$ magnesium reacts with excess sulfuric acid? $\underset{\mathrm{Magnesium}}{\mathrm{Mg}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Magnesium}\mathrm{sulfate}}{{\mathrm{MgSO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$

Answer 1

Step 1: Given data

$6\mathrm{g}$ of Magnesium

Step 2: Applying Stoichiometry

From the reaction, $\underset{\mathrm{Magnesium}}{\mathrm{Mg}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfuric}\mathrm{acid}}{{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{l}\right)}\to \underset{\mathrm{Magnesium}\mathrm{sulfate}}{{\mathrm{MgSO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrogen}\mathrm{gas}}{{\mathrm{H}}_2\left(\mathrm{g}\right)}$ we can say that

One mole of Magnesium yields one mole of Magnesium sulphate.

Number of moles of Magnesium$=$Number of moles Magnesium Sulfate

Moles of Magnesium$=\frac{\mathrm{Given}\mathrm{mass}}{\mathrm{Molecular}\mathrm{mass}}\Rightarrow \frac{6}{24}\Rightarrow 0.25\mathrm{moles}$

So, moles of Magnesium Sulfate$=0.25\mathrm{moles}$

The molecular weight of one Magnesium Sulfate$=24+32+4\times 16\Rightarrow 120\mathrm{g}$

So, the molecular weight of $0.25\mathrm{moles}$ Magnesium Sulfate$=120\times 0.25\Rightarrow 30\mathrm{g}$

Therefore, the molecular weight of Magnesium Sulfate when $6\mathrm{g}$ Magnesium reacts with Sulfuric acid is $30\mathrm{g}$