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jasenka [17]
2 years ago
11

During an experiment, 104 grams of calcium carbonate reacted with an excess amount of hydrochloric acid. If the percent yield of

the reaction was 80.15%, what was the actual amount of calcium chloride formed? CaCO3 + HCl → CaCl2 + CO2 + H2O 90.1 grams 92.4 grams 109.2 grams 115.3 grams
Chemistry
1 answer:
postnew [5]2 years ago
7 0

Answer:

92.4 grams.

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> = (104 g)/(100.08 g/mol) = <em>1.039 mol.</em>

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

1.0 mole of CaCO₃ produce → 1.0 mole of CaCl₂.

∴ 1.039 mole of CaCO₃ produce → 1.039 mole of CaCl₂.

∴ The amount of CaCl₂ produced = no. of moles x molar mass = (1.039 mol)(110.98 g/mol) = 114.3 g.

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

Percent yield of the reaction = 80.15%, theoretical yield = 115.3 g.

<em>∴ actual yield = [(percent yield of the reaction)(theoretical yield)]/100 </em>= [(80.15%)/(115.3 g)] / 100 = <em>92.42 g ≅ 92.4 g.</em>

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