Question

Que masa de metano y oxigeno se necesitan para obtener 60gr de dioxido de carbono?

Answer 1

Answer:

$\large \boxed{\text{22 g methane; 60 g oxygen}}$

Explanation:

1. Gather the information

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

Mᵣ:  16.04      32.00        44.01

        CH₄   +   2O₂   ⟶   CO₂   +   2H₂O

m/g:                                   60

 

1. Moles of CO₂

$\text{Moles of CO}_{2} = \text{60 g CO}_{2} \times \dfrac{\text{1 mol CO}_{2}}{\text{44.01 g CO}_{2}} = \text{1.36 mol CO}_{2}$

2. Mass of CH₄

(a) Moles of CH₄

$\text{Moles of CH}_{4} = \text{1.36 mol CO}_{2} \times \dfrac{\text{1 mol CH}_{4}}{\text{1 mol CO}_{2}} = \text{1.36 mol CH}_{4}$

(b) Mass of CH₄

$\text{Mass of CH}_{4} = \text{1.36 mol CH}_{4} \times \dfrac{\text{16.04 g CH}_{4}}{\text{1 mol CH}_{4}} = \textbf{22 g CH}_{4}\\\\\text{The mass of CH _{4} required is \large \boxed{\textbf{22 g}} }$

3. Mass of O₂ required

(a) Moles of O₂

$\text{Moles of O}_{2} = \text{1.36 mol CO}_{2} \times \dfrac{\text{2 mol O}_{2}}{\text{1 mol CO}_{2}} = \text{2.73 mol O}_{2}$

(b) Mass of O₂

$\text{Mass of O}_{2} = \text{2.73 mol O}_{2} \times \dfrac{\text{32.00 g O}_{2}}{\text{1 mol O}_{2}} = \textbf{60 g O}_{2}\\\\\text{The mass of O _{2} required is \large \boxed{\textbf{60 g}} }$