Question

⦁   (12 pts) Calcium carbide, CaC2, reacts with water to form calcium hydroxide and the flammable gas acetylene, C2H2. (Historically, this reaction was used for bicycle lamps and miner’s helmets!) When 96.5 g of calcium carbide was reacted with 52.3 g of water according to the reaction below, 99.3 g of calcium hydroxide was produced. What was the percent yield of this reaction?

CaC2 + 2H2O Ca(OH)2 + C2H2

Answer 1

Moles of CaC2 = mass / molar mass

= 96.5 / 64

= 1.51 moles

Moles of water = 52.3 / 18

= 2.91

CaC2 + 2H2O = Ca(OH)2 + C2H2

Now 1 mole CaC2 require 2 mole H2O

1.51 mole will require = 1.51 * 2 = 3.02 mole

But we have only 2.91 mole H2O

So H2O is limiting reagent

Now mole of Ca(OH)2 = 1/2 mole of H2O

= 1/2 * 2.91

= 1.455

Mass of Ca(OH)2

= Mole * molar mass

= 1.455 * 74

= 107.67 grams

% yield = ( oberserved mass / calculated mass )*100

= (99.3 / 107.67 )*100

= 92.23 %

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