Question

What mass of H₂O is formed when excess H₂ reacts with 64 g of O₂

Answer 1

Answer: 72 g of $H_2O$ is formed.

Explanation:

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} O_2=\frac{64}{32}=2moles$

$2H_2+O_2(g)\rightarrow 2H_2O$

As  $H_2$ is the excess reagent, $O_2$ is the limiting reagent as it limits the formation of product.

According to stoichiometry :

1 mole of $O_2$ produce=  2 moles of $H_2O$

Thus 2 moles of $O_2$ will require=$\frac{2}{1}\times 2=4moles$  of $H_2O$

Mass of $H_2O=moles\times {\text {Molar mass}}=4moles\times 18g/mol=72g$

Thus 72 g of $H_2O$ is formed.