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

Concept: This is a reversible adiabatic process (\(q=0\)). Relation: \(TV^{\gamma-1} = \text{Constant}\). 
Step 1: Setup Equation. \(T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}\). Given \(V_2 = 8V_1\) and \(T_2 = T_1/4\). \[ T_1 V_1^{\gamma-1} = \frac{T_1}{4} (8V_1)^{\gamma-1} \] 

Step 2: Solve for \(\gamma\). \[ 4 = 8^{\gamma-1} \] \[ 2^2 = (2^3)^{\gamma-1} \implies 2 = 3(\gamma-1) \implies \gamma = 5/3 \] 

Step 3: Identify Gas. \(\gamma = 5/3\) corresponds to a **Monoatomic** gas. Helium (He) is monoatomic.