Question

How many grams of H2O will be formed when 32.0 g H2 is mixed with 84.0 g of O2 and allowed to react to form water

Answer 1

94.58 g of
`H_2O`

Step-by-step explanation:

For this question we have to start with the reaction:

`H_2~+~O_2~->~H_2O`

Now, we can balance the reaction:

`2H_2~+~O_2~->~2H_2O`

We have the amount of
`H_2` and the amount of
`O_2` . Therefore we have to find the limiting reactive, for this, we have to follow a few steps.

1) Find the moles of each reactive, using the molar mass of each compound (
`H_2~=~2~g/mol~~O_2=~32~g/mol` ).

2) Divide by the coefficient of each compound in the balanced reaction ("2" for
`H_2` and "1" for
`O_2`).

Find the moles of each reactive

`32.0~g~H_2(1~mol~H_2)/(2~g~of~H_2)=15.87~mol~H_2`

`84.0~g~of~O_2(1~mol~of~O_2)/(32~g~of~O_2)=2.62~mol~of~O_2`

Divide by the coefficient

`(15.87~mol~H_2)/(2)=7.94`

`(2.62~mol~of~O_2)/(1)=2.62`

The smallest values are for
`H_2`, so hydrogen is the limiting reagent. Now, we can do the calculation for the amount of water:

`32.0~g~H_2(1~mol~H_2)/(2~g~of~H_2)(2~mol~H_2O)/(2~mol~H_2)(18~g~H_2O)/(1~mol~H_2O)=94.58~g~H_2O`

We have to remember that the molar ratio between
`H_2O` and
`H_2` is 2:2 and the molar mass of
`H_2O` is 18 g/mol.