Question

C3H8+5O2 →3CO2+4H20,

Answer 1

Answer:

6.63 grams of CO₂ can be produced.

Explanation:

1 mol of C₃H₈ reacts to 5 moles of oxygen, in a combustion reaction in order to produce 3 moles of carbon dioxide and 4 moles of water.

First of all, we need to know, the limiting reactant:

5.52 g . 1mol / 44g = 0.125 moles of C₃H₈

8.03 g . 1mol/ 32g = 0.251 moles of oxygen.

1 mol of C₃H₈ can react to 5 moles of O₂

Then, 0.125 moles, must react to (0.125 . 5) /1 = 0.625 moles.

We only have 0.251 moles of O₂ and we need 0.625 moles. Cause we do not have enough oxygen, that's the limiting reagent.

5 moles of oxygen can produce 3 moles of CO₂

0.251 moles may produce (0.251 . 3) /5 = 0.1506 moles.

We convert the moles to mass : 0.1506 mol . 44g /mol = 6.63 g

Reaction is: C₃H₈  +  5O₂ →  3CO₂  + 4H₂O