Question

Acetylene is formed by the reaction of water with calcium carbide, according to the following

equation:

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

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

Answer 1

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

22.4L of acetylene at STP = 1 mol

10.0L of acetylene at STP = x mol

x = $\dfrac{10.0L}{22.4L} mol$ = 0.446 mol

from the equation above, 1 mol of Calcium Carbide produces 1 mole of acetylene.

Therefore, 0.446 mol of Calcium Carbide produces 0.446 mol of acetylene.

0.446 mol of Calcium Carbide = 0.446 mol (26.04 g/mol) = 11.61g.

Therefore, 11.61g grams of $CaC_2$ would be needed to produce 10.0 L of ethyne.