Calculate the volume of 2 mol $d m^{- 3}$ solution of sulfuric acid required to react with 24 g of magnesium

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Solution

$M g^{+ 2} + H_{2} {S O}_{4} ⟶ {M g S O}_{4} + 2 H^{+}$ [where $M g^{+ 2} = 24 g = 1 m o l e s$ ]
By stichiometry, $1$ mole of $M g^{+ 2}$ reacts with $1$ mole of $H_{2} {S O}_{4}$
$1 \frac{m o l e}{d m^{3}}$ of ${M g}^{+ 2}$ reacts with $1 \frac{m o l e}{d m^{3}}$ of $H_{2} {S O}_{4}$
Volume of sulphuric acid required = $\frac{2}{1} = 2 d m^{3}$

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Calculate the volume of 2 mol dm−3 solution of sulfuric acid required to react with 24 g of magnesium

Mg+2+H2SO4⟶MgSO4+2H+ [where Mg+2=24g=1moles]By stichiometry, 1 mole of Mg+2 reacts with 1 mole of H2SO41moledm3 of Mg+2 reacts with 1moledm3 of H2SO4Volume of sulphuric acid required =21=2 dm3

Calculate the volume of 2 mol $d m^{- 3}$ solution of sulfuric acid required to react with 24 g of magnesium