A 12-gram mixture of methane, CH4, and ethane, C2H6, is completely combusted in

Question

A 12-gram mixture of methane, CH4, and ethane, C2H6, is completely combusted in oxygen. If the total mass of H2O produced is 23.4 grams, what was the approximate mass of the methane in the original mixture?

Explanation / Answer

ch4 + o2 ---> co2 + 2h2o c2h6 + 02 ---> 2co2 + 3h2o 16x + 30 y = 12 2x+3y=1.272 16x + 24y = 10.177 6y = 1.822 => y = 0.3 => x= 0.18 mol CH4 Mass CH4 = 2.976 grams. Explanation of equations if needed: The initial 12g mixture has methane, with 16g/mol and ethane, with 30g/mol, therefore the mixture of x mol methane and y mol ethane gives us the following equation: 16x+30y=12. The equations of the chemical reactions let us know that the double of methane mol (2x) and the triple of ethane mol (3y) gives us the number of water mol, because of the quantums of the equation. The number of mol of water is 23.4/18. I solved the equation system by multiplying the latter equation with 8 and subtracting it from the first. I found out the y, meaning number of mol for ethane, then replaced it in the simpler equation and found out the number of methane mol (x). To obtain the mass of methane, i multiplied x with 16 g/mol.

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