Complete combustion of 3g of ethane gives x × 10²² molecules of water. The value of x is __________. (Round off to the Nearest Integer). [Use : N A = 6.023 × 10²³; Atomic masses in u : C : 12.0; O : 16.0; H : 1.0]

Solution

Chemistry ma’am · 4 steps

Every hydrogen in ethane ends up in water, so one ethane gives three water molecules.

1. Moles of ethane. Molar mass of C₂H₆ is 2 × 12 + 6 × 1 = 30 g mol⁻¹. So 3 g is 3/30 = 0.1 mol.

   2C₂H₆ + 7O₂ → 4CO₂ + 6H₂O

2. Moles of water. 2 mol of ethane give 6 mol of water, so 1 mol gives 3 mol. Then 0.1 mol of ethane gives 0.1 × 3 = 0.3 mol of H₂O.
3. Count molecules. 0.3 × 6.023 × 10²³ = 1.807 × 10²³ = 18.07 × 10²² molecules.
4. So. x = 18.07, which rounds to 18.

The common slip: 6. Skipping the mole ratio and taking 0.1 mol of water instead of 0.3 mol gives 6, not 18.

The idea to keep. In combustion, go from moles of fuel to moles of product with the balanced mole ratio first. Multiply by N_A only at the last step.

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