Question

How many grams of CaC2 would be needed to produce 10.0 L (measured at STP) acetylene?

Answer 1

$CaC_{2(s)} + 2H_2O_{(l)} = Ca(OH)_{2(s)} + C_2H_{2(g)}$

10.0L of acetylene = 10/22.4 moles = 0.446 moles

from the reaction above,

1 mole of Calcium Carbide produces 1 mole of acetylene

Therefore, 0.446 moles of Calcium Carbide will produce 0.446 moles of acetylene

1 mole of Calcium Carbide = 64g

0.446 moles of Calcium Carbide = xg

x = 0.446(64)g = 28.5g

Therefore, 28.5 grams of CaC2 would be needed to produce 10.0 L acetylene