Question

Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.6 g of methane is mixed with 64.9 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

Answer:

21.6 g

Explanation:

The reaction that takes place is:

• CH₄ + 2O₂ → CO₂ + 2H₂O

First we convert the given masses of both reactants into moles, using their respective molar masses:

• 9.6 g CH₄ ÷ 16 g/mol = 0.6 mol CH₄

• 64.9 g O₂ ÷ 32 g/mol = 2.03 mol O₂

0.6 moles of CH₄ would react completely with (2 * 0.6) 1.2 moles of O₂. As there are more O₂ moles than required, O₂ is the reactant in excess and CH₄ is the limiting reactant.

Now we calculate how many moles of water are produced, using the number of moles of the limiting reactant:

• 0.6 mol CH₄ * $\frac{2molH_2O}{1molCH_4}$ = 1.2 mol H₂O

Finally we convert 1.2 moles of water into grams, using its molar mass:

• 1.2 mol * 18 g/mol = 21.6 g