Question

When 10.0 grams of CH4 reacts completely with 40.0 grams of O2 such that there are no reactants left over, 27.5 grams of carbon dioxide are formed. How many grams of water are formed? CH4+ 2O2 → CO2 + 2H2O

Answer 1

Answer:

$\boxed{\text{27.4 g CO _{2} ; 22.5 g H _{2} O}}}$

Explanation:

We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.  

M_r:      16.04    32.00  44.01   18.02

             CH₄  +  2O₂  →  CO₂ + 2H₂O

m/g:      10.0      40.0

1. Moles of CH₄

$\text{Moles of CH}_{4} = \text{10.0 g CH}_{4} \times \dfrac{\text{1 mol CH}_{4}}{\text{16.04 g CH}_{4}} = \text{0.6234 mol CH}_{4}$

2. Mass of CO₂

(i) Calculate the moles of CO₂

The molar ratio is (1 mol CO₂ /1 mol CH₄)

$\text{Moles of CO _{2} } = \text{0.6234 mol CH _4 } \times \dfrac{\text{1 mol CO _{2} }} {\text{1 mol CH _{4} }} = \text{0.6234 mol CO _{2} }$

(ii) Calculate the mass of CO₂

$\text{Mass of CO _{2} } = \text{0.6234 mol CO _{2} } \times \dfrac{\text{44.01 g CO _{2} }}{\text{1 mol CO _{2} }} = \textbf{27.4 g CO _{2} }\\\\\text{The mass of carbon dioxide formed is } \boxed{\textbf{27.4 g CO _{2} }}$

3. Mass of H₂O

(i) Calculate the moles of H₂O

The molar ratio is (2 mol H₂O /1 mol CH₄)

$\text{Moles of H _{2} O}= \text{0.6234 mol CH}_{4} \times \dfrac{\text{2 mol H _{2} O}}{\text{1 mol CH _{4} }} = \text{1.247 mol H _{2} O}$

(ii) Calculate the mass of H₂O

$\text{Mass of H _{2} O} = \text{1.247 mol H _{2} O } \times \dfrac{\text{18.02 g H _{2} O}}{\text{1 mol H _{2} O}} = \textbf{22.5 g H _{2} O}\\\\\text{The mass of water formed is } \boxed{\textbf{22.5 g H _{2} O}}$