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

The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of = 54.219 g

Number of atoms of =

Molar mass of = 98 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 3 mole of react with 2 mole of

So, 0.553 moles of react with moles of

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

Now we have to calculate the moles of

From the reaction, we conclude that

As, 3 mole of react to give 3 mole of

So, 0.553 mole of react to give 0.553 mole of

Now we have to calculate the volume of  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 volume of hydrogen gas

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