Answer:
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.
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
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