Question

Calculate the volume of $O_2$ and $273K$ required for the complete combustion of $2.64L$
of acetylene ${(C}_2{H}_2)$ at $1$ atm and $273K$. ${2C}_2{H}_2\left(g\right)+{5O}_2\left(g\right)⟶{4CO}_2\left(g\right)+{2H}_{2}O\left(l\right)$

• A. $3.6L$
• B. $1.056L$
• C. $6.6L$
• D. $10L$

Answer 1

The correct option is C $6.6L$

Solution:- (C) 6.6 L

Given volume of acetylene $=0.25{dm}^3=2.64L$

Combustion of acetylene $\left(C_2{H}_2\right)$-

$2\underset{\left(\text{acetylene}\right)}{{C_2{H}_2}_{\left(g\right)}}+5{O_2}_{\left(g\right)}⟶4{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)}$

As we know that volume of 1 mole of gas is 22.4 L.

Amount of oxygen required for the combustion of 44.8 L (2 moles) of acetylene $=(5\times 22.4)L=112L$

$\therefore$ Amount of oxygen required for the combustion of 2.64 L of methane $=\frac{112}{44.8}\times 2.64=6.6L$

Hence, 6.6 L of oxygen required for the complete combustion of 2.64 L of acetylene.