Given the eqaution CaCl 2 + K 2 CO 3 --> CaCO 3 + KCl and mass of Mass of CaCl 2

ID: 784794 • Letter: G

Question

Given the eqaution CaCl2 + K2CO3 --> CaCO3 + KCl and mass of Mass of CaCl2= 5g Mass of K2CO3= 2.5g Mass of product = 2.8g

What is the

Theoretical yield (CaCO3):

Actual yield (CaCO3):

Percent yield:

Mass of CACl2 present in original solution:

molar mass of CaCl2, K2CO3, KCl and CaCO3 are respectively 110.98, 138.21, 74.55 and 100.09 g/mol

mole of CaCl2 = 2.5 / 110.98 = 0.023 mol

mole of K2CO3 = 2.5 / 138.21 = 0.018 mol

thus K2CO3 is limiting reagent and hence 0.018 mole of CaCO3 will be formed.

mass of CaCO3 produced = 0.018 * 100.09 = 1.8 g

Theoretical yield (CaCO3) = 1.8 g

mass of KCl produced = 0.018 * 74.55 = 1.3 g

Hence theoretical yield = 1.8 g + 1.3 g = 3.1 g

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from overall theoretical yield mass fraction of CaCO3= 1.8/3.1 = 0.58

Actual yield (CaCO3) =0.58*2.8 = 1.6 g

Actual yield = 2.8 g

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Percent yield (CaCO3)= (1.6 /1.8) * 100 = 88.9% = 89% (with two significant figures)

Percent yield (overall)= (2.8 /3.1) * 100 = 90%

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mole of CaCl2 before reaction started = 2.5 / 110.98 = 0.0225 mol

mole of CaCl2 after complete reaction = 0.0225 - 0.018 = 0.0045 mol

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mass of CaCl2 before reaction started = 2.5 g

mass of CaCl2 after complete reaction = 2.5 - 0.018*110.98 = 0.50 g

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