# The hydrogen density of a mixture of hydrogen, methane and carbon monoxide (II) is 7.8. For complete combustion of one volume

**The hydrogen density of a mixture of hydrogen, methane and carbon monoxide (II) is 7.8. For complete combustion of one volume of this mixture, 1.4 volumes of oxygen are required. Determine in% the volumetric composition of the mixture**

Let there be x l of hydrogen in the mixture, and liters of carbon monoxide (II), then the volume of methane gas [1- (x + y)] l

Gas masses at standard will be equal:

m (xl H2) = 2x \ 22.4 (g)

m (y l CO) = 28 y \ 22.4 (g)

m (CH4) = 16 [1- (x + y)] \ 22.4 (g)

Since D (H2) = 7.8, mass of 22.4 liters of the mixture = 7.8 ∙ 2

Mass of 1 liter of mixture = 7.8 ∙ 2 \ 22.4 (g)

Hence

2x \ 22.4 + 28 y \ 22.4 + 16 [1- (x + y)] \ 22.4 = 7.8 ∙ 2 \ 22.4

7 x – 6y = 0.2

When a mixture of gases burns, the following reactions occur:

2H2 + O2 = 2H2O

2CO + O2 = 2CO2

CH4 + 2O2 = CO2 + 2H2O

Based on the equations, 0.5 chl of oxygen reacted with hydrogen, 0.5 chl of gas with carbon monoxide, and 2 ∙ [1- (x + y)] l of oxygen with methane, 1.4 liters in total

0.5 x + 0.5 y + 2 ∙ [1- (x + y)] = 1.4

x + y = 0.4

We solve the system of equations

x + y = 0.4

7 x – 6y = 0.2

x = 0.2 l H2

y = 0.2 l CO

1- (x + y) = 0.6 l CH4

w (H2) = 20%

w (CO) = 20% w (CH4) = 60%

Answer: w (H2) = 20%

w (CO) = 20% w (CH4) = 60%