How much carbon dioxide is released during combustion of 48 g of methane?

| January 3, 2021 | education

In order to solve the problem, you must:
1) Write the equation of the reaction and inscribe above the substances the data that are provided according to the conditions of the problem:
CH4 + 2O2 → CO2 + 2H2O
2) Since, according to the condition of the problem, we are given the mass of methane, we need to find the amount of matter (CH4):
To find it, also using the periodic table, we find the molar mass of methane:
M (CH4) = 12 + 1 * 4 = 12 + 4 = 16 g / mol
It turns out:
ν (CH4) = m / M = 48g / 16g / mol = 3 mol
3) According to the mf reaction equation, we see that the amount of carbon dioxide is equal to the amount of methane:
ν (CO2) = ν (CH4) = 3 mol
4) Find the volume of carbon dioxide that will be released during the combustion of 48 g of methane:
V (CO2) = ν * Vm = 3 mol * 22.4 l / mol = 67.2 l.
Answer: 67.2 liters is the volume of carbon dioxide that will be released during the combustion of 48 g of methane.



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