For this question, the first thing we need to do is set up the properly balanced equation for the combustion of methane:
CH4 + 2 O2 -> CO2 + 2 H2O
We have:
2.90*10^-3 or 0.0029 grams of CH4
To find how much water was produced, we need to know how many moles of CH4 we have, we will do that by using the provided mass and its molar mass, 16.04g/mol
16.04g = 1 mol
0.0029g = x moles
x = 1.81*10^-4 moles
Now, according to the molar ratio between CH4 and H2O, we need 1 mol of CH4 to produce 2 moles of H2O, therefore we have a molar ratio of 1:2, so we will have 1.81*10^-4 * 2 = 3.61*10^-4 moles of H2O being produced
Now using the molar mass of water, 18g/mol, we can find the final mass of water:
18g = 1 mol
x grams = 3.61*10^-4 moles
x = 0.0065 grams or 6.5*10^-3 grams
For the second part of the question is the same, we know that we have 1.81*10^-4 moles of CH4, and because of the molar ratio, we will also have 3.61*10^-4 moles of O2, using the molar mass of O2, 32g/mol, we will have:
32g = 1 mol
x grams = 3.61*10^-4
x = 0.0115 grams of O2 or 1.15*10^-2 grams