Percent Yield Problem!!!?
What is the percent yield if 155 g of calcium carbonate is treated with 250g of anhydrous hydrogen chloride and only 142 g of CaCl2 is obtained?
- white rabbitLv 48 years agoFavorite Answer
First you need the molar masses of HCl, CaCl2, CaCO3
HCl: 36.46 g/mol
CaCl2: 110.98 g/mol
CaCO3: 100.09 g/mol
The equation for the reaction is:
CaCO3(s) + 2 HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)
First convert 155g of calcium carbonate into moles.
155g CaCO3 / 100.09 g/mol = 1.5486 mol CaCO3
Then do the same for your HCl
250g HCl / 36.46 g/mol = 6.8568 mol HCl
Then apply them using the equation to find the limiting reactant.
1.5486 mol CaCO3 x 1 mol CaCl2 per 1 mol CaCO3 = 1.5486 mol CaCl2
6.8568 mol HCl x 1 mol CaCl2 per 2 mol HCl = 3.4284 mol CaCl2
Since CaCO3 produces the lower value, it is the limiting reactant, and the HCl is in excess. Therefore, we know the theoretical yield of CaCl2 is 1.5486 moles. Now we convert this into grams.
1.5486 mol CaCl2 x 110.98 g/mol = 171.86g CaCl2
The percentage yield is simply (actual yield / theoretical yield) x 100
(142 / 171.86) x 100