Question

For Balanced chemical reaction 
\(\text{2Al$_{(s)}$ + 6HCl$_{(aq)}$ $\rightarrow$ 2AlCl$_3$ + 3H$_2$(g)} \)
\(\text{which of the following is correct?} \)

• A. 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.
• B. 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.
• C. With excess of HCl, moles of AlCl$_3$ produced per mole of Al reacted are 2.
• D. At given P and T, 12 L HCl produce 6 L H$_2$ gas.

Hint:

In stoichiometric calculations, always use the molar ratios from the balanced equation to find the volumes or amounts of products and reactants.

Answer 1

The Correct Option is B

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.