Answer to Question #298486 in Chemistry for Bilal saleh

Question

What is the mass of technical calcium carbide with a mass fraction of 12% required for the production of acetylene with a volume of 6.72 l (nu)?

Answer 1

Solution:

calcium carbide = CaC₂

acetylene = C₂H₂

At STP, one mole of any gas occupies a volume of 22.4 L.

Thus, 6.72 liters of C₂H₂ correspond to:

Moles of C₂H₂ = (6.72 L C₂H₂) × (1 mol C₂H₂ / 22.4 L C₂H₂) = 0.3 mol C₂H₂

Balanced chemical equation:

CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂

According to stoichiometry:

1 mol of CaC₂ produces 1 mol of C₂H₂

Thus, X mol of CaC₂ produces 0.3 mol of C₂H₂

Moles of CaC₂ = X = (0.3 mol C₂H₂) × (1 mol CaC₂ / 1 mol C₂H₂) = 0.3 mol CaC₂

The molar mass of CaC₂ is 64.1 g/mol

Hence,

Mass of CaC₂ = (0.3 mol CaC₂) × (64.1 g CaC₂ / 1 mol CaC₂) = 19.23 g CaC₂

Calculate the mass of technical CaC₂:

Mass of technical CaC₂ = (19.23 g CaC₂) × (100% / 12%) = 160.25 g

Answer: The mass of technical calcium carbide (CaC₂) is 160.25 grams