Question

For the following reaction, calculate how many moles of each product are formed when 4.05 g of water is used. 2H20 →2H2 + O2

Answer 1

0.45 grams of hydrogen and 3.53 grams of oxygen will be formed, when 4.05 grams of water is used.

Step-by-step explanation:

The given reaction is
`2H_20 \rightarrow 2H_2 + O_2`

2 moles of
`H_2O` is used to form 2 moles of
`H_2` and 1 mole of
`O_2`

As 1 mole of
`H_2O=18 g`, 1 mole of
`H_2=2 g`, and 1 mole of
`O_2=32 g`.

So, 36 grams of
`H_2O` is used to form 4 grams of
`H_2` and 32 grams of
`O_2`

Therefore, 1 gram of
`H_2O` is used to form 4/36 grams of
`H_2` and 32/36 grams of
`O_2`

Therefore, 4.05 gram of
`H_2O` is used to form (1/9)4.09=0.45 grams of
`H_2` and (8/9)4.05=3.53 grams of
`O_2`.

Hence, 0.45 grams of hydrogen and 3.53 grams of oxygen will be formed, when 4.05 grams of water is used.