What volume of acetylene can be obtained from technical calcium carbide weighing 65 g
What volume of acetylene can be obtained from technical calcium carbide weighing 65 g, if the mass fraction of impurities in it is 20%?
The reaction for producing acetylene from calcium carbide occurs in accordance with the following chemical reaction equation:
CaC2 + 2H2O = C2H2 + Ca (OH) 2;
The decomposition of 1 mol of calcium carbide releases 1 mol of gaseous acetylene.
The mass of pure calcium carbide is 65 x 0.8 = 52 grams.
Let’s find the amount of a substance contained in 52 grams of calcium carbide.
M CaC2 = 40 + 12 x 2 = 64 grams / mol;
N CaC2 = 52/64 = 0.8 mol;
When this reaction is carried out, 0.8 mol of acetylene will be released.
Let’s calculate its volume.
One mole of gas under normal conditions fills a volume of 22.4 liters.
The amount of acetylene will be equal to:
V C2H2 = 0.8x 22.4 = 17.92 liters;