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Answer :
D. When 1.21 g of zinc reacts with 50.0 mL of 0.120 mol/L hydrochloric acid, 74.4 mL of hydrogen gas is produced at SATP.
To determine the volume of hydrogen gas produced when 1.21 g of zinc reacts with 50.0 mL of 0.120 mol/L hydrochloric acid at SATP, follow these steps:
- Write the balanced chemical equation:
The reaction between zinc and hydrochloric acid is given by:
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)
- Calculate moles of zinc:
The molar mass of zinc (Zn) is 65.38 g/mol. Therefore, the moles of zinc are:
n(Zn) = (1.21 g) / (65.38 g/mol)
≈ 0.0185 mol Zn
- Calculate moles of hydrochloric acid (HCl):
The concentration of HCl is 0.120 mol/L and the volume is 50.0 mL (or 0.050 L). Therefore, the moles of HCl are:
n(HCl) = 0.120 mol/L × 0.050 L
= 0.006 mol HCl
- Determine the limiting reactant:
The stoichiometry of the reaction requires 2 moles of HCl for every mole of Zn. Therefore, the limiting reactant is HCl:
0.006 mol HCl / 2 = 0.003 mol Zn needed, which is less than 0.0185 mol Zn available.
- Calculate moles of hydrogen gas (H₂) produced:
According to the stoichiometry, 1 mole of Zn produces 1 mole of H₂:
n(H₂) = 0.006 mol HCl / 2
= 0.003 mol H₂
- Calculate the volume of hydrogen gas at SATP:
At Standard Ambient Temperature and Pressure (SATP), 1 mole of gas occupies 24.8 L.
V(H₂) = 0.003 mol H₂ × 24.8 L/mol
≈ 0.0744 L or 74.4 mL
The correct answer is (D) 74.4 mL.
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