Question

How many grams of CO2 are produced by the combustion of 289 g of a mixture that is 28.6% CH4 and 71.4% C3H8 by mass?

Answer 1

Answer:

$\large \boxed{\text{845 g}}$

Explanation:

You have two combustion reactions — one of methane and one of propane.

For each component, we must calculate the mass, then do the conversions

mass of X ⟶ moles of X ⟶ moles of CO₂

Then, we can find the total moles of CO₂ and convert to grams.

1. Mass of each component

(a) Methane

$\text{Mass of CH}_{4} = \text{289 g mixture} \times \dfrac{\text{28.6 g CH}_{4}}{\text{ 100 g mixture}} = \text{82.65 g CH}_{4}$

(b) Propane

$\text{Mass of C _{3} H}_{8} = \text{289 g mixture} \times \dfrac{\text{71.4 g C _{3} H}_{8}}{\text{ 100 g mixture}} = \text{206.3 g C _{3} H}_{8}$

2. CO₂ from methane

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       16.04               44.01

            CH₄ + 2O₂ ⟶ CO₂ +2H₂O

m/g:    82.65

You start with the substance for which they give you numbers.

For example, this question gives you the masses of CH₄ and C₃H₈ and asks you to find the total mass of CO₂.  

(a) Moles of CH₄

$\text{Moles of CH}_{4} = \text{82.65 g CH}_{4}\times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}}= \text{5.153 mol CH}_{4}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} = \text{5.153 mol CH}_{4} \times \dfrac{\text{1 mol CO}_{2}}{\text{1 mol CH}_{4}} = \text{5.153 mol CO}_{2}$

3. CO₂ from propane

Mᵣ:       44.10                44.01

           C₃H₈ + 5O₂ ⟶ 3CO₂ +4H₂O

m/g:    206.3

(a) Moles of C₃H₈

$\text{Moles of C _{3} H}_{8} = \text{206.3 g C _{3} H}_{8}\times \dfrac{\text{1 mol C _{3} H}_{8}}{\text{44.10 g C _{3} H}_{8}}= \text{4.679 mol C _{3} H}_{8}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} = \text{4.679 mol C _{3} H}_{8}} \times \dfrac{\text{3 mol CO}_{2}}{\text{1 mol C _{3} H}_{8}} = \text{14.04 mol CO}_{2}$

4. Mass of CO₂

(a) Total moles of CO₂

Total moles = 5.153 + 14.04 = 19.19 mol CO₂

(b) Mass of CO₂

$\text{Mass of CO}_{2} = \text{19.19 mol CO}_{2} \times \dfrac{\text{44.01 g CO}_{2}}{\text{1 mol CO}_{2}} = \textbf{845 g CO}_{2}\\\text{The mass of CO _{2} produced is \large \boxed{\textbf{845 g}} }$