100 mL of a mixture of methane and ethane requires 260 mL of oxygen for complete combustion (all volumes at the same temperature and pressure). Find the volume of $CO X_{2}$ formed in mL, to the nearest integer.

Numerical answer type, answer in mL, tolerance 1.

Answer: 140 mL

CH X_{4} + 2 O X_{2} CO X_{2} + 2 H X_{2} O

and

C X_{2} H X_{6} + 3.5 O X_{2} 2 CO X_{2} + 3 H X_{2} O

. Let ethane be y mL: 2(100 - y) + 3.5y = 260, giving y = 40 mL and methane 60 mL.

CO X_{2}

= 60 + 2(40) = 140 mL.