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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