What volume of nitrogen gas at STP would react with 37.2 g of magnesium to produce magnesium nitride

The volume of nitrogen gas is 11.58 L of N₂

Mole = mass /Molar mass

= 37.2 / 24

= 1.55 moles

Now

1.55 moles of Mg will react with

= (1.55 × 1)/3

= 0.517 mole of N₂

Thus, 0.517 moles of N₂ is needed for the reaction.

Now

1 mole of a gas occupies 22.4 L at STP.

So,

Nitrogen gas is

= 0.517 × 22.4

= 11.58 L at STP

Learn more about the gas here: link

11.58 L of N₂

Explanation:

We'll begin by calculating the number of mole in 37.2 g of magnesium. This can be obtained as follow:

Mass of Mg = 37.2 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 37.2 / 24

Mole of Mg = 1.55 moles

Next, we shall write the balanced equation for the reaction. This is illustrated below:

3Mg + N₂ —> Mg₃N₂

From the balanced equation above,

3 moles of Mg reacted with 1 mole of N₂.

Therefore, 1.55 moles of Mg will react with = (1.55 × 1)/3 = 0.517 mole of N₂

Thus, 0.517 mole of N₂ is need for the reaction.

Finally, we shall determine the volume of N₂ needed for the reaction as follow:

Recall:

1 mole of a gas occupies 22.4 L at STP.

1 mole of N₂ occupied 22.4 L at STP.

Therefore, 0.517 mole of N₂ will occupy = 0.517 × 22.4 = 11.58 L at STP

Thus, 11.58 L of N₂ is needed for the reaction.

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