When 8.74 g \( \text{MnO}_2 \) is treated with HCl, then what will be the weight of \( \text{Cl}_2(g) \) obtained?
Molar mass of \( \text{MnO}_2 = 87.4 \, \text{g/mol} \).
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When performing stoichiometric calculations, ensure to use the mole ratio from the balanced equation to convert between reactants and products.
Step 1: Write the chemical reaction.
The reaction between manganese dioxide (\( \text{MnO}_2 \)) and hydrochloric acid (HCl) produces chlorine gas (\( \text{Cl}_2 \)) according to the following balanced equation:
\[
\text{MnO}_2 + 4 \, \text{HCl} \rightarrow \text{MnCl}_2 + 2 \, \text{H}_2\text{O} + \text{Cl}_2
\]
Step 2: Use stoichiometry.
From the balanced equation, 1 mole of \( \text{MnO}_2 \) produces 1 mole of \( \text{Cl}_2 \).
Step 3: Calculate the moles of \( \text{MnO}_2 \).
The molar mass of \( \text{MnO}_2 \) is 87.4 g/mol. So, the moles of \( \text{MnO}_2 \) in 8.74 g is:
\[
\text{moles of } \text{MnO}_2 = \frac{8.74 \, \text{g}}{87.4 \, \text{g/mol}} = 0.1 \, \text{mol}
\]
Step 4: Calculate the moles of \( \text{Cl}_2 \).
Since 1 mole of \( \text{MnO}_2 \) produces 1 mole of \( \text{Cl}_2 \), the moles of \( \text{Cl}_2 \) produced is also 0.1 mol.
Step 5: Calculate the mass of \( \text{Cl}_2 \).
The molar mass of chlorine gas (\( \text{Cl}_2 \)) is 70.9 g/mol. Therefore, the mass of \( \text{Cl}_2 \) produced is:
\[
\text{mass of } \text{Cl}_2 = 0.1 \, \text{mol} \times 70.9 \, \text{g/mol} = 7.1 \, \text{g}
\]
Thus, the weight of \( \text{Cl}_2 \) obtained is 7.1 g.
