Answer:
[tex]\large \boxed{\text{0.453 g of H$_{2}$; 3.60 g of O$_{2}$}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
MM: 2.016 32.00
2H₂O ⟶ 2H₂ + O₂
m/g: 4.05
1. Mass of hydrogen
(a) Moles of H₂O
[tex]\text{Moles of H$_{2}$O} = \text{4.05 g H$_{2}$O}\times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O }}= \text{0.2248 mol H$_{2}$O}[/tex]
(b) Moles of H₂
[tex]\text{Moles of H$_{2}$} = \text{0.2248 mol H$_{2}$O } \times \dfrac{\text{2 mol H$_{2}$}}{\text{2 mol H$_{2}$O }} = \text{0.2248 mol H$_{2}$}[/tex]
(c) Mass of H₂
[tex]\text{Mass of H$_{2}$} =\text{0.2248 mol H$_{2}$} \times \dfrac{\text{2.016 g H$_{2}$}}{\text{1 mol H$_{2}$}} = \textbf{0.453 g H$_{2}$}\\\\\text{The reaction produces $\large \boxed{\textbf{0.453 g}}$ of H$_{2}$}[/tex]
2. Mass of oxygen
(a) Moles of O₂
[tex]\text{Moles of O$_{2}$} = \text{0.2248 mol H$_{2}$O } \times \dfrac{\text{1 mol O$_{2}$}}{\text{2 mol H$_{2}$O}} = \text{0.1124 mol O$_{2}$}[/tex]
(b) Mass of O₂
[tex]\text{Mass of O$_{2}$} =\text{0.1124 mol O$_{2}$} \times \dfrac{\text{32.00 g O$_{2}$}}{\text{1 mol O$_{2}$}} = \textbf{3.60 g O$_{2}$}\\\\\text{The reaction produces $\large \boxed{\textbf{0.453 g of H$_{2}$ and 3.60 g of O$_{2}$}}$}[/tex]
