Magnesium nitride can be produced by passing nitrogen gas over magnesium metal.

Question

3Mg+N2 --> Mg3N2

IF 28.8g of Mg are mixed with 21g of N2 and allowed to react, which is the limiting reagent?

Answer 1

I solved it now. Thanks

Answer 2

Calculate the number of moles of Mg and N2 available. There need to be at least three times as many moles of Mg is N2 is the limiting reagent. If there are less than three times as many moles of Mg, then Mg is the limiting reagent.

We will be glad to critique your work.

Answer 3

When TWO reactants are given, you know it is a limiting reagent problem. Here is how you do these.
1. Convert what you are given to mols.
a. 28.8 g Mg to mols Mg. # mols = grams/atomic mass Mg.
b. 21 g N2 to mols N2. # mols = grams/atomic mass N2.

2. Using the coefficients in the balanced equation, convert mols Mg AND mols N2 (separately) to mols of Mg3N2.
a. mols Mg3N2 = mols Mg x (1 mol Mg3N2/3 mols Mg) = ??
b. mols Mg3N2 = mols N2 x (1 mol Mg3N2/1 mol N2) = ??
c. The one producing the lower number of mols of product (Mg3N2) is the limiting reagent.

Answer 4

The moles of each reactant are:
(8.8g of Mg) / (24.3gMg/moleMg) = 0.362 moles Mg
21.0g N2 / 28.01 g/mol N2 = 0.750 mol N2

The mole ratio of Mg to N2 in the chemical equation is 3/1

The mole ratio in the mixture is
0.362 mol Mg / 0.750 mol N2 or
0.483 mol Mg / 1 mol N2 ( a LOT less than 3 to 1)
That means not enough Mg to react with all of the N2 present. That also means Mg is the limiting reagent.