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What mass of zinc will be required to react completely with 100 cm3 of 0.1 moldm⁻³ hydrochloric acid? Determine volume of hydrogen gas that would be evolved at s.t.p.

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2 parts
1

0.325 g\text{0.325 g}

2

112 cm3\text{112 cm}^3

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Here is the solution for Question 8:

Part 8

What mass of zinc will be required to react completely with 100 cm3100 \text{ cm}^3 of 0.1 moldm−30.1 \text{ moldm}^{-3} hydrochloric acid? Determine volume of hydrogen gas that would be evolved at s.t.p.
Equation of reaction: Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)\text{Zn}_{(s)} + 2\text{HCl}_{(aq)} \longrightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}
[Zn = 65, molar volume = 22.4 dm3mol−122.4 \text{ dm}^3\text{mol}^{-1}]

  • Given:
    • Volume of HCl=100 cm3=0.1 dm3\text{HCl} = 100 \text{ cm}^3 = 0.1 \text{ dm}^3
    • Concentration of HCl=0.1 moldm−3\text{HCl} = 0.1 \text{ moldm}^{-3}
    • Molar mass of Zn=65 g/mol\text{Zn} = 65 \text{ g/mol}
    • Molar volume at s.t.p. = 22.4 dm3mol−122.4 \text{ dm}^3\text{mol}^{-1}
    • Equation: Zn(s)+2HCl(aq)⟶ZnCl2(aq)+H2(g)\text{Zn}_{(s)} + 2\text{HCl}_{(aq)} \longrightarrow \text{ZnCl}_{2(aq)} + \text{H}_{2(g)}

Part 1

0.325 g\text{0.325 g}

Calculate the mass of zinc required.

  1. Calculate the moles of hydrochloric acid (HCl\text{HCl}).

    Moles of HCl=Concentration×Volume\text{Moles of HCl} = \text{Concentration} \times \text{Volume} Moles of HCl=0.1 moldm−3×0.1 dm3\text{Moles of HCl} = 0.1 \text{ moldm}^{-3} \times 0.1 \text{ dm}^3 Moles of HCl=0.01 mol\text{Moles of HCl} = 0.01 \text{ mol}

  2. Determine the moles of zinc (Zn\text{Zn}) required using the mole ratio from the balanced equation.

    From the equation, 1 mol1 \text{ mol} of Zn\text{Zn} reacts with 2 mol2 \text{ mol} of HCl\text{HCl}.

    Moles of Zn=Moles of HCl×1 mol Zn2 mol HCl\text{Moles of Zn} = \text{Moles of HCl} \times \frac{1 \text{ mol Zn}}{2 \text{ mol HCl}} Moles of Zn=0.01 mol×12\text{Moles of Zn} = 0.01 \text{ mol} \times \frac{1}{2} Moles of Zn=0.005 mol\text{Moles of Zn} = 0.005 \text{ mol}

  3. Calculate the mass of zinc required.

    Mass of Zn=Moles of Zn×Molar mass of Zn\text{Mass of Zn} = \text{Moles of Zn} \times \text{Molar mass of Zn} Mass of Zn=0.005 mol×65 g/mol\text{Mass of Zn} = 0.005 \text{ mol} \times 65 \text{ g/mol} Mass of Zn=0.325 g\text{Mass of Zn} = \boxed{\text{0.325 g}}

Part 2

112 cm3\text{112 cm}^3

Determine the volume of hydrogen gas evolved at s.t.p.

  1. Determine the moles of hydrogen gas (H2\text{H}_2) produced using the mole ratio from the balanced equation.

    From the equation, 2 mol2 \text{ mol} of HCl\text{HCl} produces 1 mol1 \text{ mol} of H2\text{H}_2.

    Moles of H2=Moles of HCl×1 mol H22 mol HCl\text{Moles of H}_2 = \text{Moles of HCl} \times \frac{1 \text{ mol H}_2}{2 \text{ mol HCl}} Moles of H2=0.01 mol×12\text{Moles of H}_2 = 0.01 \text{ mol} \times \frac{1}{2} Moles of H2=0.005 mol\text{Moles of H}_2 = 0.005 \text{ mol}

  2. Calculate the volume of hydrogen gas evolved at s.t.p.

    Volume of H2=Moles of H2×Molar volume at s.t.p.\text{Volume of H}_2 = \text{Moles of H}_2 \times \text{Molar volume at s.t.p.} Volume of H2=0.005 mol×22.4 dm3mol−1\text{Volume of H}_2 = 0.005 \text{ mol} \times 22.4 \text{ dm}^3\text{mol}^{-1} Volume of H2=0.112 dm3\text{Volume of H}_2 = 0.112 \text{ dm}^3

    Convert to cm3\text{cm}^3:

    Volume of H2=0.112 dm3×1000 cm3/dm3\text{Volume of H}_2 = 0.112 \text{ dm}^3 \times 1000 \text{ cm}^3/\text{dm}^3 Volume of H2=112 cm3\text{Volume of H}_2 = \boxed{\text{112 cm}^3}
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