Gaseous methane CH4 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose 0.80 g of methane is mixed with 1.92 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.

1.08 g of water

Explanation:

The balanced chemical equation for the reaction of combustion of methane (CH₄) is the following:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

According to the equation, 1 mol of CH₄ reacts with 2 moles of O₂. We convert from mol to grams by using the molar masses:

1 mol CH₄ = (1 x 12 g/mol) + (4 x 1 g/mol) = 16 g

2 mol O₂ = 2 x (2 x 16 g/mol) = 64 g

2 mol H₂O = 2 x ((2 x 1 g/mol) + 16 g/mol)= 36 g

From the masses of reactants (CH₄ and O₂), we can see that the stoichiometric ratio is 64 g O₂/16 g CH₄ = 4.

First, we have to identify which reactant is the limiting reactant. We can compare the stoichiometric ratio with the actual reactants ratio (the masses of reactants we have):

1.92 g O₂/0.80 g CH₄ = 2.4

As 4>2.4, we can conclude that O₂ is the limiting reactant.

Now, we consider the stoichiometric ratio between the limiting reactant (64 g O₂) and the product we have to calculate (36 g H₂O), and we multiply the ratio by the actual mass of O₂:

1.92 g O₂ x 36 g H₂O/64 g O₂ = 1.08 g

Therefore, 1.08 g of H₂O will be produced by the chemical reaction of 0.80 g of methane with 1.92 g of oxygen.

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