For Balanced chemical reaction \(\text{2Al$_{(s)}$ + 6HCl$_{(aq)}$ $\rightarrow$ 2AlCl$_3$ + 3H$_2$(g)} \) \(\text{which of the following is correct?} \)
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In stoichiometric calculations, always use the molar ratios from the balanced equation to find the volumes or amounts of products and reactants.
With excess of Al, volume of H$_2$ gas produced per mole of HCl reacted will be 33.6 L at 1 atm & 273 K.
With excess of Al, volume of H$_2$ gas produced per mole of HCl reacted will be 11.2 L at 1 atm & 273 K.
With excess of HCl, moles of AlCl$_3$ produced per mole of Al reacted are 2.
At given P and T, 12 L HCl produce 6 L H$_2$ gas.
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The Correct Option isB
Solution and Explanation
Step 1: Analyzing the reaction.
The balanced chemical equation shows that 2 moles of Al react with 6 moles of HCl to produce 3 moles of H$_2$. Therefore, the moles of H$_2$ produced per mole of HCl reacted are:
\[
\frac{3}{6} = \frac{1}{2} \text{ mole of H}_2 \text{ per mole of HCl}.
\]
Step 2: Volume of H$_2$ gas produced.
Using the molar volume of an ideal gas (22.4 L at 1 atm and 273 K), the volume of H$_2$ gas produced per mole of HCl reacted will be:
\[
\text{Volume of H}_2 = \frac{1}{2} \times 22.4 = 11.2 \, \text{L}.
\]
Step 3: Conclusion.
Thus, the correct answer is (2), as 11.2 L of H$_2$ gas are produced per mole of HCl reacted at 1 atm and 273 K.
