What mass of oxygen is needed for the complete combustion of 5.80×10−3 g of methane?

First, we need to balance the stoichiometry equation.

?CH4(g) + ?O2(g) --> ?CO2(g) + ?H2O(g)

Lets start by putting 1 for each question mark. There are 2 hydrogens in water, and 4 in methane. This means whatever number we put in front of methane, the number in front of H2O has to be double that. If we put a 2 in front of H2O, the only thing unbalanced in the equation in the number of oxygens: there are four on the right side and only two in the oxygen on the left. We can make the coefficient in front of oxygen 2:

1CH4(g) + 2O2(g) --> 1CO2(g) + 2H2O(g)

But, to use the stoichiometry equation, we need to figure out the number of moles of methane used, instead of grams. The molecular weight of methane is 16.043g/mol. We can use this to find the moles of methane:

1.2*10^-3g / (16.043g/mol)=7.48*10^-5 moles of methane

Now that we know the number of moles of methane combusted, we can use the stoichiometry equation to find out how many moles of CO2 this equates to. For every mole of methane burned, a mole of CO2 is produced:

7.48*10^-5 mol CH4 * (1 mol CO2 / 1 mol CH4) = 7.48*10^-5 mol CO2

We know that the reaction goes to completion, so all the methane will be reacted. To find the mass of carbon dioxide used, we can use the number of moles from the above equation, and the molecular weight of CO2 (= 44.009 g/mol).

7.48*10^-5 mol CO2 * 44.009 g/mol = 3.29*10^-3 g CO2

The mass of CO2 produced is 3.29*10^-3g.