$\underset{4 \mathrm{gm}}{\mathrm{CH}_4+2 \mathrm{O}_2} \longrightarrow \underset{11 \mathrm{gm}}{\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}}$
$0.25 \text { mole } \quad 0.25 \text { mole }$
$\left.0.25 \text { mole } \mathrm{CH}_4 \text { gives } 0.25 \text { mole (or } 11 \mathrm{gm}\right) \mathrm{CO}_2$
$CaC_2 + H_2O → Ca(OH)_2 + C_2H_2$