Magnesium reacts with nitrogen according to the following reaction: Mg + N2 Mg3N

Question

Magnesium reacts with nitrogen according to the following reaction: Mg + N2 Mg3N2

(a) Name the product of this reaction

(b) Balance the reaction equation

(c) What is reduced and what is oxidized in this reaction?

(d) How many grams of Mg3N2 can be prepared from 24.3 g of magnesium and excess nitrogen?

(e) Magnesium-copper alloy contains 80.0% magnesium by mass; copper does not react with nitrogen. How many grams of this alloy are needed to prepare 10.1 g of Mg3N2?

(f) How many grams of Mg3N2 will be obtained in the reaction between 2.43 g of magnesium and 1.40 g of nitrogen?

Explanation / Answer

3Mg + N2 -------------> Mg3N2

(a) Magnesium nitride ( Mg3N2)

(b) 3Mg + N2 -------------> Mg3N2

(c) Nitrogen is reduced and Magnesium is oxidised

N (0) --------------> N(3-)

Mg(0) ---------------> Mg(2+)

oxidation states are indicated in paranthesis.

(d) moles Mg = mass / molar mass = 24.3 / 24 = 1.01 mol

According to balanced equation, 3 moles Mg reats with N2 to produce 1 mole of Mg3N2 .

1 mole Mg3N2 = 100 g

amount of Mg3N2 formed from 1.01 moles of Mg = 1,01 x (1/3) x 100 = 33.7 g

(e) Amount of Mg required to form 10.1 g of Mg3N2 = 10.1 x 72/100 = 7.27 g

But 80% of Mg only present in the alloy.

So, the amount of alloy required = 7.27 x 100 / 80 = 9.09 g

(f)

moles of Mg = 2.43 / 24 = 0.101 mol

moles of N2 = 1.40 / 28 = 0.05

Therefore , Mg is the limiting reactant.

Amount of Mg3N2 can be produced = 0.101 x ( 1/3 ) x 100 = 3.37 g

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