Question

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

`0.45gH_(2)`

`3.6gO_(2)`

Step-by-step explanation:

The problem gives you the balanced reaction:

`_(2)H_(2)O=_(2)H_(2)+_(2)O_(2)`

To calculate the mass formed of each product you need to have the molar mass of reactant and products, so:

molar mass of
`H_(2)` =
`2(g)/(mol)`

molar mass of
`O_(2)` =
`32(g)/(mol)`

molar mass of
`H_(2)O` =
`18(g)/(mol)`

Then you should use the stoichiometry to make the relationships between the moles of products and the reactant:

- For
`H_(2)`:

`4.05gH_(2)O*(1molH_(2)O)/(18gH_(2)O)*(2molesH_(2))/(2molesH_(2)O)*(2gH_(2))/(1molH_(2))=0.45gH_(2)`

-For
`O_(2)`:

`4.05gH_(2)O*(1molH_(2)O)/(18gH_(2)O)*(1molO_(2))/(2molesH_(2)O)*(32gO_(2))/(1molO_(2))=3.6gO_(2)`