Answer:
11.08 g
Explanation:
The reaction given is:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
First, we need to know which reactant is limiting. So let's do the stoichiometry calculus between methane and oxygen, testing methane as the limiting reactant.
By the balanced equation, 1 mol of CH₄ reacts with 2 moles of O₂. The molar masses are CH₄: 16 g/mol, O₂: 32 g/mol
1*16 g/mol of CH₄ ---------------------- 2*32 g/mol of O₂
8.7 g ----------------------- x
By a simple direct three rule:
16x = 556.8
x = 34.8 g
So, it's needed 34.8 g of oxygen to react with 8.7 g of methane. Because there is only 19.7 g of oxygen, it is limiting, and methane is in excess.
Now we can do the stoichiometry calculus between the limiting reactant and the water, that has 18 g/mol as molar mass. By the reaction, 2 moles of O₂ form 2 moles of H₂O, so we can use 1:1
1 * 32 g/mol of O₂ ---------------------- 1*18 g/mol of H₂O
19.7 g of O₂ ------------------------------ y
By a simple direct three rule:
32y = 354.6
y = 11.08 g
