Answer:
[tex]\boxed{\text{27.4 g CO$_{2}$; 22.5 g H$_{2}$O}}}[/tex]
Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place. Â
M_r: Â Â Â 16.04 Â Â 32.00 Â 44.01 Â 18.02
       CH₄  +  2O₂  →  CO₂ + 2H₂O
m/g: Â Â Â 10.0 Â Â Â 40.0
1. Moles of CHâ‚„
[tex]\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}[/tex]
2. Mass of COâ‚‚
(i) Calculate the moles of COâ‚‚
The molar ratio is (1 mol COâ‚‚ /1 mol CHâ‚„)
[tex]\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$_{2}$}[/tex]
(ii) Calculate the mass of COâ‚‚
[tex]\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$_{2}$}}[/tex]
3. Mass of Hâ‚‚O
(i) Calculate the moles of Hâ‚‚O
The molar ratio is (2 mol Hâ‚‚O /1 mol CHâ‚„)
[tex]\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}[/tex]
(ii) Calculate the mass of Hâ‚‚O
[tex]\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}}[/tex]
