Question

Given the following reaction Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) If 0.500 g of Ca reacts with 0.850 g HCl and 1.21 g CaCl2 are produced, what is the theoretical yield, the percent yield and the limiting reactant?

1.38 g, 87.7%, Ca limiting

1.38 g, 93.8%, HCl limiting

1.29 g, 87.7%, Ca limiting

1.29 g, 93.8% HCl limiting

1.38 g, 92.8%, Ca limiting

1.29 g, 92.8%, HCl limiting

Answer 1

Moles of Ca = Given Weight of Ca / Molar Mass of Ca = 0.5/40.08 = 0.0124 moles ,(Molar Mass of Ca is 40.08g/mol)

Moles of HCl = Given Weight of HCl / Molar Mass of HCl = 0.85/36.46 = 0.0233 moles ,(Molar Mass of HCl is 36.46g/mol)

Moles of CaCl2 (produced practically) = Given Weight of CaCl2 / Molar Mass of CaCl2 = 1.21 / 110.98 = 0.0109 moles, (Molar Mass of Ca Cl2 is 110.98 g/mol)

Intial Final Ca(s) + 2HCl + Ca Cly(aq) + Hz (9) 0.0124 0.0233 0.0124-00 0.0233-200 oC 3C • Sf Ga(s) is Limiting Reagent thero 0.0124 - 2C = 0 => OG = 0.0124 • ff Hce is Limiting Reagent then 0.0233-23C = 0 => OC = 0.01165 The value from which we get least value of ac is the limiting Reagent Therefore He is the Limiting Reagent. Moles of cace, (the oretical) = uc = 0.01165 moles Mass of Gace2Cthecretical) = Moles X Molar Maso of Cacéz = 0.01165 X 110.98 = 1.299 Pos.cont Yild - / Actual Yield x 100 ( Theoretical Yield ) 1.2 1. xloo - 93.8 % 1.29

Theoretical Yield of CaCl2 = 1.29g, HCl is the limiting reagent, and precent yield is 93.8%.