What volume of acetylene measured at standard conditions could be obtained from 10 grams of calcium chloride?

Answer 1

The reaction is :
CaC2 + 2 H2O >> C2H2 + Ca(OH)2
Molecular weight CaC2 = 64 g/mol
10 g / 64 = 0.156 mole of calcium carbide ( not calcium chloride )
The ratio between CaC2 and C2H2 is 1 : 1so we would get 0.156 mole of C2H2
At STP : T = 273 K and p = 1 atm
V = nRT / p = 0.156 x 0.0821 x 273 / 1 = 3.5 L of C2H2

Answer 2

You mean, of course, calcium carbide, CaC2.

CaC2 + 2H2O ----> C2H2 + Ca(OH)2

Now you can work out the volume yourself.