Question

CaC2 (s) + 2H2O (l) LaTeX: \longrightarrow
⟶
C2H2 (g) + Ca(OH)2 (aq)

Given the reaction above, what is the maximum amount (in liters) of C2H2 that could be produced if 160 g of CaC2 is reacted in 100 g of H2O?

Answer 1

Moles of CaC2;

$Moles = \dfrac{mass}{RFM}$

RFM of CaC2 = 40 + (12x2) = 64

$Moles = \dfrac{160g}{64g/mol} = 2.5mol$

Moles of H2O $= \dfrac{100g}{18g/mol} = 5.56mol$

Moles of water that would react with 2.5 mol of CaC2

Mole ratio of CaC2 : H2O = 1:2

Therefore, Moles of H2O $= \dfrac{2}{1} x 2,5mol = 5mol$

Yet there were 5.56 mol of H2O. Therefore, CaC2 if the limiting reagent

Moles of C2H2 that would be produced by 2.5 mol of CaC2;

Mole ratio of CaC2 : C2H2 = 1:1

Thus mole so C2H2 $= \dfrac{1}{1}x2.5mol = 2.5mol$

1 mol of a gas at RTP = 24L

2.5 mol = ?

$= \dfrac{2.5mol x 24L}{1mol} = 60L of C2H2 at RTP$

Volume of C2H2 produced would be 60 Litres at RTP