Question

Lpg is a mixture of n-butane and iso-butane. The volume of oxygen needed to burn 1 kg of lpg at ntpwould be

Answer 1

Both iso-butane and n- butane are the isomers with same molecular formula , and the reaction proceeds as follows-
2C₄H₁₀+13O₂→8CO₂+10H₂O
this equation shows that 2 mol of mixture reacts with 13 mol of O_{2
​​​​​​}molecular wt. of butane= 2$\times$ 58 = 116g
therefore, 116 g of butane requires 13 $\times$ 22.4L of oxygen
so, 1kg g of LPG is burnt by = 13*22.4*1000 / 116 = 2510.34