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=(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=(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* \text{ Molar mass of }H_2O`

Molar mass of
`H_2O` = 18 g/mole

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

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