Answer to Question #161541 in Chemistry for Sah

Question

C2H2 can be produced by the reaction of calcium carbide with water according to the following equation

CaC2(s)+ H2O(I) = C2H2 (g)+Ca(OH)2(aq) if 48.6g of calcium carbide is reacted at STP, what is the volume of the acetylene produced?

Answer 1

Solution:

The balanced chemical equation:

CaC₂(s) + 2H₂O(I) = C₂H₂(g) + Ca(OH)₂(aq)

According to the equation above: n(CaC₂) = n(C₂H₂)

Moles of CaC₂ = Mass of CaC₂ / Molar mass of CaC₂

The molar mass of CaC₂ is 64.099 g mol^-1.

Hence,

n(CaC₂) = 48.6 g / 64.099 g mol^-1 = 0.7582 mol

n(C₂H₂) = n(CaC₂) = 0.7582 mol

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

Thus 0.7582 mol of C₂H₂ occupies:

(0.7582 mol × 22.4 L) / 1 mol = 16.98 L of C₂H₂

V(C₂H₂) = 16.98 L = 17.0 L

Answer: 17.0 L of the acetylene (C₂H₂) produced.