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General ChemistryHard

Stoichiometry — limiting reactant

React 10.0 g Zn with 20.0 mL of 6.00 M HCl.
Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g).
M(Zn) = 65.38 g/mol. Determine limiting reactant and volume of H₂ produced (STP, 22.4 L/mol).
Given data
m(Zn) = 10.0 gV(HCl) = 20.0 mL = 0.0200 L[HCl] = 6.00 MM(Zn) = 65.38 g/molV<sub>m</sub> (STP) = 22.4 L/mol
Review the theory: Chimica Generale
Steps
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  1. Calculate moles of Zn: n(Zn) = 10.0/65.38.
  2. Calculate moles of HCl: n(HCl) = M·V = 6.00 × 0.0200.
  3. Required HCl = 2·n(Zn) = 2 × 0.1529.
  4. Which is the limiting reactant?
  5. Moles of H₂ produced: n(H₂) = n(HCl)/2.
  6. Volume H₂ at STP: V = n·22.4 L/mol.
Full worked solution
  1. Calculate moles of Zn: n(Zn) = 10.0/65.38.
    10.0/65.3810.0/65.38
    n(Zn) = 0.1529 mol.
  2. Calculate moles of HCl: n(HCl) = M·V = 6.00 × 0.0200.
    6.00⋅0.02006.00 \cdot 0.0200
    n(HCl) = 0.120 mol.
  3. Required HCl = 2·n(Zn) = 2 × 0.1529.
    2⋅0.15292 \cdot 0.1529
    Need 0.3058 mol HCl, have only 0.120 mol.
  4. Which is the limiting reactant?
    HCl is the limiting reactant.
  5. Moles of H₂ produced: n(H₂) = n(HCl)/2.
    0.120/20.120/2
    n(H₂) = 0.120/2 = 0.0600 mol.
  6. Volume H₂ at STP: V = n·22.4 L/mol.
    0.0600⋅22.40.0600 \cdot 22.4
    V(H₂) = 0.0600 × 22.4 = 1.344 L.
Result:Limiting reactant: HCl — V(H₂) produced = 1.34 L at STP.
TheoryChimica GeneraleFormula sheetChemistry

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