# Question #73939

Oct 21, 2014

2.3 L of butane will be burned.

${C}_{4} {H}_{10} + 6.5 {O}_{2} \rightarrow 4 C {O}_{2} + 5 {H}_{2} O$

So 6.5 mol ${O}_{2}$ will burn 1 mol of ${C}_{4} {H}_{10}$

Since Avogadro said that equal volumes of gases contain the same number of moles we can write:

6.5 L of ${O}_{2}$ will burn 1L of ${C}_{4} {H}_{10}$

So 1L of ${O}_{2}$ will burn 1/6.5 L of ${C}_{4} {H}_{10}$

So 15 L of ${O}_{2}$ will burn (1/6.5) x 15 = 2.3 L