Question

How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phosphoric acid (H3PO4)

The equation is 3Mg + 2H3(PO4)-->Mg(PO4)2+3H2

Answer 1

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of
`H_3PO_4` = 54.219 g

Number of atoms of
`Mg` =
`7.179* 10^(23)`

Molar mass of
`H_3PO_4` = 98 g/mol

First we have to calculate the moles of
`H_3PO_4` and
`Mg`.

`\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}`

`\text{Moles of }H_3PO_4=(54.219g)/(98g/mol)=0.553mol`

and,

`\text{Moles of }Mg=(7.179* 10^(23))/(6.022* 10^(23))=1.19mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2`

From the balanced reaction we conclude that

As, 3 mole of
`Mg` react with 2 mole of
`H_3PO_4`

So, 0.553 moles of
`Mg` react with
`(2)/(3)* 0.553=0.369` moles of
`H_3PO_4`

From this we conclude that,
`H_3PO_4` is an excess reagent because the given moles are greater than the required moles and
`Mg` is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of
`H_2`

From the reaction, we conclude that

As, 3 mole of
`Mg` react to give 3 mole of
`H_2`

So, 0.553 mole of
`Mg` react to give 0.553 mole of
`H_2`

Now we have to calculate the volume of
`H_2` gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies
`0.553* 22.4=12.4L` volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L