The mass of magnesium required to produce 220 mL of hydrogen gas at STP on reaction with excess of dilute HCl is: (Given molar mass of Mg = 24 g/mol)
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In gas-related stoichiometric problems, always remember to use the molar volume of gases at STP (22.4 L or 22400 mL) to convert between volume and moles.
The balanced equation for the reaction between magnesium and hydrochloric acid is:
\[
\text{Mg} (s) + 2 \text{HCl} (aq) \rightarrow \text{MgCl}_2 (aq) + \text{H}_2 (g)
\]
The stoichiometry indicates a 1:1 molar ratio between Mg and H\(_2\). Given a hydrogen gas volume of 220 mL at STP, we convert this to moles.
1. At STP, 1 mole of an ideal gas occupies 22.4 L (22400 mL). The moles of hydrogen gas are calculated as:
\[
n_{\text{H}_2} = \frac{220}{22400} = 0.00982 \, \text{mol}
\]
2. Due to the 1:1 molar ratio, the moles of Mg required are equal to the moles of H\(_2\), which is 0.00982 mol.
3. The mass of magnesium is then determined using its molar mass:
\[
\text{Mass of Mg} = n_{\text{Mg}} \times \text{Molar mass of Mg} = 0.00982 \times 24 = 0.44 \, \text{g}
\]
Therefore, 0.44 g of magnesium is required.
