Question

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 8.7 g of methane is mixed with 19.7 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to significant digits.

Answer 1

11.08 g

Step-by-step 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