Answer to Question #153580 in General Chemistry for John

Question #153580

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?




1
Expert's answer
2021-01-04T03:48:17-0500

"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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