Answer:
[tex]0.45gH_{2}[/tex]
[tex]3.6gO_{2}[/tex]
Explanation:
The problem gives you the balanced reaction:
[tex]_{2}H_{2}O=_{2}H_{2}+_{2}O_{2}[/tex]
To calculate the mass formed of each product you need to have the molar mass of reactant and products, so:
molar mass of [tex]H_{2}[/tex] = [tex]2\frac{g}{mol}[/tex]
molar mass of [tex]O_{2}[/tex] = [tex]32\frac{g}{mol}[/tex]
molar mass of [tex]H_{2}O[/tex] = [tex]18\frac{g}{mol}[/tex]
Then you should use the stoichiometry to make the relationships between the moles of products and the reactant:
- For [tex]H_{2}[/tex]:
[tex]4.05gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{2molesH_{2}}{2molesH_{2}O}*\frac{2gH_{2}}{1molH_{2}}=0.45gH_{2}[/tex]
-For [tex]O_{2}[/tex]:
[tex]4.05gH_{2}O*\frac{1molH_{2}O}{18gH_{2}O}*\frac{1molO_{2}}{2molesH_{2}O}*\frac{32gO_{2}}{1molO_{2}}=3.6gO_{2}[/tex]
