100 g of propane is completely reacted with 1000 g of oxygen. The mole fraction of carbon dioxide in the resulting mixture is x × 10⁻². The value of x is ____________. (Nearest integer)
[Atomic weight : H = 1.008; C = 12.00; O = 16.00]

Solution

Chemistry ma’am · 3 steps

Propane runs out first, so the leftover oxygen stays in the mixture and counts in the total.

1. Balanced equation. Each mole of propane needs 5 mol of oxygen and gives 3 mol of CO₂ and 4 mol of H₂O.

   C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

2. Find the moles. The molar mass of C₃H₈ is 3 × 12.00 + 8 × 1.008 = 44.064 g mol⁻¹, so 100 g is 100/44.064 = 2.269 mol. Oxygen is 1000/32 = 31.25 mol. Propane needs only 5 × 2.269 = 11.347 mol of O₂, so propane is limiting and oxygen is in excess.

   Species	Initial (mol)	Change (mol)	Final (mol)
   C₃H₈	2.269	-2.269	0
   O₂	31.25	-11.347	19.903
   CO₂	0	+3 × 2.269	6.808
   H₂O	0	+4 × 2.269	9.078
   Total	33.519	+2.269	35.789

3. So. The mole fraction of CO₂ is 6.808/35.789 = 0.190 = 19.0 × 10⁻², so x is 19.

The common slip: 43. Leaving out the unreacted oxygen makes the total only 15.886 mol, which gives 43 instead of 19.

The idea to keep. For a mole fraction after a reaction, list every species present at the end: all the products and the unused excess reagent. The total moles must include all of them.

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