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julia-pushkina [17]
3 years ago
6

During an experiment, 102 grams of calcium carbonate react with an excess amount of hydrochloric acid. If the percent yield of t

he reaction was 85.15%, what was the amount of calcium chloride formed?​
Chemistry
1 answer:
Nata [24]3 years ago
8 0

Answer:

96.31 g.

Explanation:

  • From the balanced reaction:

<em>CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O,</em>

1.0 mole of CaCO₃ reacts with 2.0 moles of HCl to produce 1.0 mole of CaCl₂, 1.0 mole of CO₂, and 1.0 mole of H₂O.

  • We need to calculate the no. of moles of (104 g) of CaCO₃:

<em>no. of moles of CaCO₃ = mass/molar mass </em>= (102 g)/(100.08 g/mol) = <em>1.019 mol.</em>

<u><em>Using cross multiplication:</em></u>

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.019 mole of CaCO₃ produce → 1.019 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.019 mol)(110.98 g/mol) = 113.1 g.

∵ percent yield of the reaction = [(actual yield)/(theoretical yield)] x 100.

Percent yield of the reaction = 85.15%, theoretical yield = 113.1 g.

<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100</em> = [(85.15%)(113.1 g)] / 100 = <em>96.31 g.</em>

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