Question

Suppose 4.0 g of hydrogen reacts completely with 32.0 g of oxygen to form one product what is the mass of the product?

Answer 1

Answer: The mass of product, $H_2O$ is, 36.0 grams.

Explanation : Given,

Mass of $H_2$ = 4.0 g

Mass of $O_2$ = 32.0 g

Molar mass of $H_2$ = 2 g/mol

Molar mass of $O_2$ = 32 g/mol

First we have to calculate the moles of $H_2$ and $O_2$.

$\text{Moles of }H_2=\frac{\text{Given mass }H_2}{\text{Molar mass }H_2}$

$\text{Moles of }H_2=\frac{4.0g}{2g/mol}=2.0mol$

and,

$\text{Moles of }O_2=\frac{\text{Given mass }O_2}{\text{Molar mass }O_2}$

$\text{Moles of }O_2=\frac{32.0g}{32g/mol}=1.0mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$2H_2+O_2\rightarrow 2H_2O$

From the balanced reaction we conclude that

2 mole of $H_2$ react with 1 mole of $O_2$

From this we conclude that, there is no limiting and excess reagent.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

2 moles of $H_2$ react to give 2 moles of $H_2O$

Now we have to calculate the mass of $H_2O$

$\text{ Mass of }H_2O=\text{ Moles of }H_2O\times \text{ Molar mass of }H_2O$

Molar mass of $H_2O$ = 18 g/mole

$\text{ Mass of }H_2O=(2.0moles)\times (18g/mole)=36.0g$

Therefore, the mass of product, $H_2O$ is, 36.0 grams.