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A) What observations are made during and after the reaction.

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2 parts
c

0.08 mol\text{0.08 mol}

d

10.88 g\text{10.88 g}

Step-by-step solution

9 steps

Part a

What observations are made during and after the reaction.

  • During heating, the white solid turns yellow and a colorless gas is produced.
  • After cooling, the yellow solid turns back to white.

Part b

Give a balanced equation for the reaction.

ZnCO3(s)→heatZnO(s)+CO2(g)\text{ZnCO}_3(\text{s}) \xrightarrow{\text{heat}} \text{ZnO}(\text{s}) + \text{CO}_2(\text{g})

Part c

0.08 mol\text{0.08 mol}

Calculate the moles of CO2\text{CO}_2 formed in the reaction.

  1. Calculate the mass of CO2\text{CO}_2 formed.

    The mass lost during heating is due to the formation of CO2\text{CO}_2 gas.

    Mass of CO2=Initial mass of ZnCO3−Mass of residue (ZnO)\text{Mass of CO}_2 = \text{Initial mass of ZnCO}_3 - \text{Mass of residue (ZnO)} Mass of CO2=10g−6.48g=3.52g\text{Mass of CO}_2 = 10\text{g} - 6.48\text{g} = 3.52\text{g}

  2. Calculate the molar mass of CO2\text{CO}_2.

    Molar mass of CO2=C+2×O=12+2×16=44g/mol\text{Molar mass of CO}_2 = \text{C} + 2 \times \text{O} = 12 + 2 \times 16 = 44\text{g/mol}

  3. Calculate the moles of CO2\text{CO}_2.

    Moles of CO2=Mass of CO2Molar mass of CO2=3.52g44g/mol=0.08mol\text{Moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} = \frac{3.52\text{g}}{44\text{g/mol}} = 0.08\text{mol}

    The moles of CO2\text{CO}_2 formed are 0.08 mol\boxed{\text{0.08 mol}}.

Part d

10.88 g\text{10.88 g}

If the residue was reacted with excess hydrochloric acid, calculate the mass of zinc chloride formed.

  1. Write the balanced equation for the reaction of the residue (ZnO\text{ZnO}) with hydrochloric acid.

    ZnO(s)+2HCl(aq)→ZnCl2(aq)+H2O(l)\text{ZnO}(\text{s}) + 2\text{HCl}(\text{aq}) \to \text{ZnCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l})

  2. Calculate the molar mass of ZnO\text{ZnO}.

    Molar mass of ZnO=Zn+O=65+16=81g/mol\text{Molar mass of ZnO} = \text{Zn} + \text{O} = 65 + 16 = 81\text{g/mol}

  3. Calculate the moles of ZnO\text{ZnO} in the residue.

    Moles of ZnO=Mass of ZnOMolar mass of ZnO=6.48g81g/mol=0.08mol\text{Moles of ZnO} = \frac{\text{Mass of ZnO}}{\text{Molar mass of ZnO}} = \frac{6.48\text{g}}{81\text{g/mol}} = 0.08\text{mol}

  4. Determine the moles of ZnCl2\text{ZnCl}_2 formed.

    From the balanced equation, the mole ratio of ZnO\text{ZnO} to ZnCl2\text{ZnCl}_2 is 1:11:1.
    Therefore, Moles of ZnCl2=Moles of ZnO=0.08mol\text{Moles of ZnCl}_2 = \text{Moles of ZnO} = 0.08\text{mol}.

  5. Calculate the molar mass of ZnCl2\text{ZnCl}_2.

    Molar mass of ZnCl2=Zn+2×Cl=65+2×35.5=65+71=136g/mol\text{Molar mass of ZnCl}_2 = \text{Zn} + 2 \times \text{Cl} = 65 + 2 \times 35.5 = 65 + 71 = 136\text{g/mol}

  6. Calculate the mass of ZnCl2\text{ZnCl}_2 formed.

    Mass of ZnCl2=Moles of ZnCl2×Molar mass of ZnCl2\text{Mass of ZnCl}_2 = \text{Moles of ZnCl}_2 \times \text{Molar mass of ZnCl}_2 Mass of ZnCl2=0.08mol×136g/mol=10.88g\text{Mass of ZnCl}_2 = 0.08\text{mol} \times 136\text{g/mol} = 10.88\text{g}

    The mass of zinc chloride formed is 10.88 g\boxed{\text{10.88 g}}.

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