Question

3. For the following reaction, calculate how many grams of each product are formed when 4.05 g

of water is used.
2 H2O → 2 H2 + O2

Answer 1

`\large \boxed{\text{0.453 g of H$_(2)$; 3.60 g of O$_(2)$}}`

Step-by-step 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

`\text{Moles of H$_(2)$O} = \text{4.05 g H$_(2)$O}* \frac{\text{1 mol H$_(2)$O}}{\text{18.02 g H$_(2)$O }}= \text{0.2248 mol H$_(2)$O}`

(b) Moles of H₂

`\text{Moles of H$_(2)$} = \text{0.2248 mol H$_(2)$O } * \frac{\text{2 mol H$_(2)$}}{\text{2 mol H$_(2)$O }} = \text{0.2248 mol H$_(2)$}`

(c) Mass of H₂

`\text{Mass of H$_(2)$} =\text{0.2248 mol H$_(2)$} * \frac{\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$_(2)$}`

2. Mass of oxygen

(a) Moles of O₂

`\text{Moles of O$_(2)$} = \text{0.2248 mol H$_(2)$O } * \frac{\text{1 mol O$_(2)$}}{\text{2 mol H$_(2)$O}} = \text{0.1124 mol O$_(2)$}`

(b) Mass of O₂

`\text{Mass of O$_(2)$} =\text{0.1124 mol O$_(2)$} * \frac{\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$_(2)$}}$}`