Burning 15 liters of a mixture of CH4 and C2H6 produced 27 liters of carbon dioxide. How many С2Н6 molecules are there per СН4 molecule in the initial mixture?
CH4 + 2O2 = CO2 + 2H2O
2С2Н6 + 7О2 = 4СО2 + 6Н2О
The reaction equations show that one molecule of carbon dioxide is formed from one molecule of methane, and there are already two molecules of CO2 for one molecule of ethane.
Let x be the volume of methane, y the volume of ethane.
x + y = 15 (according to the condition of the problem, a mixture of 15 liters was taken)
The volume of CO2 generated by combustion of x liters of methane is x.
The volume of CO2 generated by the combustion of liters of ethane is 2u.
According to the condition, the volume of released CO2 is 27 liters, then x + 2y = 27.
Let’s compose a system of equations:
x + y = 15,
x + 2y = 27;
y = 15 – x,
x + 2 * (15 – x) = 27;
y = 15 – x,
x + 30 – 2x = 27;
y = 12,
x = 3.
There are 12 ethane molecules per 3 methane molecules, or 4 ethane molecules per 1 methane molecule.
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