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

Answer:

94.58 g of $H_2O$

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

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}=15.87~mol~H_2$

$84.0~g~of~O_2\frac{1~mol~of~O_2}{32~g~of~O_2}=2.62~mol~of~O_2$

Divide by the coefficient

$\frac{15.87~mol~H_2}{2}=7.94$

$\frac{2.62~mol~of~O_2}{1}=2.62$

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

$32.0~g~H_2\frac{1~mol~H_2}{2~g~of~H_2}\frac{2~mol~H_2O}{2~mol~H_2}\frac{18~g~H_2O}{1~mol~H_2O}=94.58~g~H_2O$

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.