A 12-gram mixture of methane, CH4, and ethane, C2H6, is completely combusted in oxygen. If the total mass of H2O produced is 22.9 grams, what was the approximate mass of the methane in the original mixture?
I don't understand how to do this?
I started it with
22.9 g h20 * 1 mol h20/18 g h20 = 1.2722 mol
But I have no idea where to go from there or if I did that right?
I don't understand how to do this?
I started it with
22.9 g h20 * 1 mol h20/18 g h20 = 1.2722 mol
But I have no idea where to go from there or if I did that right?
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This is a hard question, so don't feel bad.
You calculated 1.2722 mol H2O. This is, you will note 2.5444 mol of H atoms.
Your 12-g mixture of methane and ethane has, let's say, x moles CH4 and y moles C2H6. We have two variables; fortunately, we have two relations between the variables. We know that there are 2.5444 mol H atoms total, i.e. 4x + 6y = 2.5444. You see where that comes from? Each mole of CH4 contributes 4 moles H, and each mole of C2H6 contributes 6 moles H to the combustion mixture. The other relationship is that the mass of methane, which is about 16x (i.e. x moles times 16 g/mol) plus the mass of ehtane, which is about 30y (y moles times 30 g/mol) must = 12 g. So we have the equation 16x + 30 y = 12.
So now you have an algebra problem: 4x + 6y = 2.5444 and 16x + 30y = 12.
There is more than one way to solve the problem, but one is to multiply 4x + 6y = 2.5444 by -4, getting -16x + -24 y = -10.1776 and add it to 16x + 30 y = 12, getting 6y = 11.8224 and so y - 0.3037 or approximately y = 0.3 moles ethane. Now 16x + 30y = 16x + 9 = 12 ---> 16x = 3 ---> x = 0.1875 moles methane.
Finally, 0.1875 mole * 16 g/mol = 3 g of methane. The other 9 g of the mixture is ethane.
You calculated 1.2722 mol H2O. This is, you will note 2.5444 mol of H atoms.
Your 12-g mixture of methane and ethane has, let's say, x moles CH4 and y moles C2H6. We have two variables; fortunately, we have two relations between the variables. We know that there are 2.5444 mol H atoms total, i.e. 4x + 6y = 2.5444. You see where that comes from? Each mole of CH4 contributes 4 moles H, and each mole of C2H6 contributes 6 moles H to the combustion mixture. The other relationship is that the mass of methane, which is about 16x (i.e. x moles times 16 g/mol) plus the mass of ehtane, which is about 30y (y moles times 30 g/mol) must = 12 g. So we have the equation 16x + 30 y = 12.
So now you have an algebra problem: 4x + 6y = 2.5444 and 16x + 30y = 12.
There is more than one way to solve the problem, but one is to multiply 4x + 6y = 2.5444 by -4, getting -16x + -24 y = -10.1776 and add it to 16x + 30 y = 12, getting 6y = 11.8224 and so y - 0.3037 or approximately y = 0.3 moles ethane. Now 16x + 30y = 16x + 9 = 12 ---> 16x = 3 ---> x = 0.1875 moles methane.
Finally, 0.1875 mole * 16 g/mol = 3 g of methane. The other 9 g of the mixture is ethane.