In this question, we have a situation in which we have to find the limiting and excess reactant in a stoichiometry problem, first thing we have to do is to set up the properly balanced reaction:
CH4 + 2 O2 -> CO2 + 2 H2O
Now we can see that the molar ratios are:
1 CH4 = 2 O2
1 CH4 = 2 H2O
2 O2 = 2 H2O
To find the limiting and excess reactants, we need to find the number of moles of both CH4 and O2, let's start with CH4, which has a molar mass of 16.04g/mol
16.04g = 1 mol
4.87g = x moles
x = 0.304 moles of CH4
If we have 0.304 moles of CH4, according to the molar ratio, we need 2 times that of O2, 0.304 * 2 = 0.608 moles of O2, but we need to check if we have this amount of moles or not, using its molar mass, 32g/mol
32g = 1 mol
11.58g = x moles
x = 0.362 moles of O2, this means that we have less O2 than it is needed, therefore O2 is the limiting reactant and CH4 is in excess
Now to find the final mass of H2O, we need to use the number of moles of the limiting reactant, 0.362 moles, and according to the molar ratio, we have the same number of moles of H2O, 0.362 moles, now using its molar mass, 18g/mol, we can find the final mass:
18g = 1 mol
x grams = 0.362 moles
x = 6.52 grams of H2O are produced