To determine the limiting reagent, find the moles of each reactant and their stoichiometry in the reaction.
Step 1: Calculate moles of Zinc (Zn)
The molar mass of zinc (Zn) is approximately 65.38 g/mol.
\[ \text{Moles of Zn} = \frac{\text{mass}}{\text{molar mass}} = \frac{32.5 , \text{g}}{65.38 , \text{g/mol}} \approx 0.497 , \text{mol} \]
Step 2: Calculate moles of Hydrochloric Acid (HCl)
The concentration of hydrochloric acid is 2 mol/dm³, and we have 200 cm³ (or 0.2 dm³).
\[ \text{Moles of HCl} = \text{concentration} \times \text{volume} = 2 , \text{mol/dm}^3 \times 0.2 , \text{dm}^3 = 0.4 , \text{mol} \]
Step 3: Write and balance the reaction equation
The balanced chemical equation for the reaction of zinc with hydrochloric acid is:
\[ \text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2 \]
From the equation, 1 mole of Zn reacts with 2 moles of HCl.
Step 4: Determine the limiting reagent
The reaction shows that:
- 1 mole of Zn reacts with 2 moles of HCl.
From the moles calculated:
- Moles of Zn: 0.497 mol
- Moles of HCl: 0.4 mol
To completely react with 0.497 moles of Zn, the required moles of HCl would be:
\[ \text{Required moles of HCl} = 0.497 , \text{mol Zn} \times 2 \approx 0.994 , \text{mol HCl} \]
Since we only have 0.4 moles of HCl, HCl is the limiting reagent.
Step 5: Calculate the excess reagent (Zn) left in moles
From the reaction stoichiometry:
- 0.4 moles of HCl would react with:
\[ \text{Moles of Zn used} = \frac{0.4 , \text{mol HCl}}{2} = 0.2 , \text{mol Zn} \]
Remaining moles of Zn:
\[ \text{Moles of Zn remaining} = 0.497 , \text{mol Zn} - 0.2 , \text{mol Zn} = 0.297 , \text{mol Zn} \]
Step 6: Calculate the volume of hydrogen gas (H₂) liberated at STP
From the balanced equation:
- 1 mole of Zn produces 1 mole of H₂
Thus, 0.4 moles of HCl will yield:
\[ \text{Moles of H₂} = \frac{0.4 , \text{mol HCl}}{2} = 0.2 , \text{mol H₂} \]
At standard temperature and pressure (STP), 1 mole of gas occupies a volume of 22.4 liters (or 22400 cm³).
The volume of H₂ gas is:
\[ \text{Volume of H₂} = 0.2 , \text{mol} \times 22400 , \text{cm}^3/\text{mol} = 4480 , \text{cm}^3 \]
Summary of Results
- Limiting Reagent: HCl
- Excess Reagent (Zn) remaining: 0.297 mol
- Volume of H₂ liberated at STP: 4480 cm³