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%

Remember: The process of learning a person lasts a lifetime. The value of the same knowledge for different people may be different, it is determined by their individual characteristics and needs. Therefore, knowledge is always needed at any age and position.