When 128 g of calcium carbide interacted with water weighing 73 g, acetylene with a volume of 40 liters was released.
When 128 g of calcium carbide interacted with water weighing 73 g, acetylene with a volume of 40 liters was released. Determine the acetylene yield as a percentage of the theoretically possible.
When calcium carbide interacts with water, acetylene and calcium hydroxide are synthesized. The reaction is described by the following equation.
CaC2 + 2H2O = Ca (OH) 2 + C2H2;
From 1 mole of calcium carbide, 1 mole of acetylene is synthesized.
Let’s calculate the available amount of calcium carbide substance.
To do this, we divide the mass of the substance by its molar weight.
M CaC2 = 40 + 12 x 2 = 64 grams / mol;
N CaC2 = 128/64 = 2 mol;
The amount of water is:
M H2O = 2 + 16 = 18 grams / mol;
N H2O = 73/18 = 4.056 mol;
Let’s calculate the volume of 2 mol of acetylene.
To do this, multiply the amount of substance by the volume of 1 mole, which is 22.4 liters.
V C2H2 = 2 x 22.4 = 44.8 liters;
The reaction yield will be K = 40 / 44.8 = 0.893 = 89.3%;