Question

# A procedure calls for 8.25 g of CaCl2 to be dissolved in water and reacted with...

A procedure calls for 8.25 g of CaCl2 to be dissolved in water and reacted with 5.00 g of K2CO3. How many grams of CaCO3 can theoretically be made? If 3.25 g of CaCO3 were actually made, what is the percent yield?

CaCl2 + K2CO3       →    2KCl + CaCO3

Calculate the formula weight for each reactant:

Ca = 40.08 g/mol

+    2 Cl = 70.90 g/mol

= 110.98 g/mol CaCl2

2 K = 78.20 g/mol

3 O = 48.00 g/mol

+          C = 12.01 g/mol

= 138.20 g/mol K2CO3

mass of CaCl2 = 8.25 g

moles of CaCl2 = 8.25 g / 110.98 g/mol ==0.0743 mol

mass of K2CO3 = 5.0 g

moles of K2CO3 = 5.0 g / 138.20 g/mol = 0.0362 mol

1 mole of CaCl2 reacts with 1 mole of K2CO3 .

here we have less moles of K2CO3 than CaCl2

so, K2CO3 is limiting reactant. from 1:1 ratio of reaction, 1 mole of CaCO3 is formed

So moles of CaCO3 formed = 0.036 moles

Mass of CaCO3 = 0.0362 moles x 100 g/mol = 3.62 grams

actual CaCO3 made = 3.25 g

percentage yield = (3.25 / 3.62 )x 100 = 90%

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