"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.
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