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Thirty points to who ever can answer this problem

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

A. 90.1 grams
B. 92.4 grams
C. 109.2 grams
D. 115.3 grams

Respuesta :

Answer:

B. 92.4 g

Explanation:

the balanced equation for the reaction is as follows

CaCO₃ + 2HCl ---> CaCl₂ + CO₂ + H₂O

molar ratio of CaCO₃ to CaCl₂ is 1:1

number of CaCO₃ moles reacted - 104 g / 100 g/mol = 1.04 mol

therefore number of CaCl₂ moles reacted - 1.04 mol

mass of CaCl₂ expected to be formed = 1.04 mol x 111 g/mol = 115.44 g

percentage yield = actual yield / theoretical yield x 100 %

theoretical yield = 115.44 g

percentage yield = 80.15 %

substituting these values in the above equation

80.15 % = actual yield / 115.44 g x 100 %

actual yield = 92.5 g

therefore answer is B. 92.4 g

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