Answer:
4 g
Explanation:
First you need to write a balanced chemical equation. You are given thatmethane is burned, meaning a combustion reaction in which carbon dioxide and water are released.
Unbalanced: CH4 + O2 ---> CO2 + H2O
Balanced: CH4 + 2O2 ---> CO2 + 2H2O
Givens:
X grams CH4 (Molecular mass 16.0 grams)
9 grams H2O (Molecular mass 18.0 grams)
11 grams CO2 (Molecular mass 44.0 grams)
Mole ratio: 1:2:1:2 (CH4:O2:CO2:H2O)
Then you need to find which of the reactants are the limiting reactant and which is the excess reactant. The limiting reactant is what you will base the rest of the problem on. To do this, you convert each measurement to moles from grams.
11 g CO2 / (44.0 g) = 0.25 moles CO2
9 g H2O / (18.0 g) = 0.5 moles
n of CH4 = n of CO2 = n of H2O /2 = 0.25 moles
m of CH4 = n* Mw = 0.25 * 16.0 = 4 g