The moles of methane required to produce 81 g of water after complete combustion is _____________ × 10⁻² mol. [nearest integer]

Solution

Chemistry ma’am · 4 steps

Turn the water into moles first, then let the 1 : 2 ratio in the balanced equation do the rest.

1. Balance the combustion. Methane burns completely to carbon dioxide and water. One carbon and four hydrogens give one CO₂ and two H₂O.

   CH₄ + 2O₂ → CO₂ + 2H₂O

2. Moles of water. The molar mass of H₂O is 18 g mol⁻¹, so 81 g is 81/18 = 4.5 mol.
3. Use the ratio. Each mole of CH₄ gives 2 moles of H₂O. So the methane needed is 4.5/2 = 2.25 mol.
4. So. 2.25 mol = 225 × 10⁻² mol, so the answer is 225.

The common slip: 450. Taking the moles of methane equal to the moles of water, and forgetting the 2 in front of H₂O, gives 4.5 mol, that is 450, not 225.

The idea to keep. In stoichiometry, always go grams to moles, then use the balanced equation's coefficients, then convert back if needed. Never compare grams directly.

Answer from the NTA final answer key. Solution by PrepWiser.