Question

5. Given 18.5 grams of CHA and 24.0 grams of Oz the following reaction occurs. Calculate the number of grams of

Answer 1

Answer:

16.5 g of CO₂ are produced by 18.5 of methane and 24 g of O₂

Explanation:

This is a reaction of combustion:

CH₄ +  2O₂  →  CO₂  + 2H₂O

1 mol of methane react to 2 moles of oxygen in order to produce 1 mol of carbon dioxide and 2 moles of water.

First of all, we determine the moles of each reactant:

18.5 g . 1mol/ 16g = 1.15 moles of methane

24 g . 1mol / 32g = 0.75 moles of oxygen

Now, we determine the limiting reactant:

1 mol of methane reacts to 2 moles of oyxgen

1.15 moles of methane may react to (1.15 . 2) /1 = 2.3 moles.

We only have 0.75 moles of O₂ and we need 2.3, that's why the oxygen is the limiting reagent, because we do not have enough oxygen for the reaction.

2 moles of O₂ produce 1 mol of CO₂

Then 0.75 moles may produce (0.75 . 1) /2 = 0.375 moles of CO₂

We convert the moles to mass → 0.375 mol . 44g /mol = 16.5 g