Magnesium nitride can be produced by passing nitrogen gas over magnesium metal.
3Mg+N2 --> Mg3N2
IF 28.8g of Mg are mixed with 21g of N2 and allowed to react, which is the limiting reagent?
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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.
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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.
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.
To determine the limiting reagent, we need to compare the stoichiometry of the reactants and see which one is fully consumed first.
First, we need to convert the given masses of Mg and N2 into moles. To do that, we use the molar masses of the elements:
Molar mass of Mg = 24.31 g/mol
Molar mass of N2 = 28.02 g/mol
Moles of Mg = mass / molar mass = 28.8 g / 24.31 g/mol = 1.186 mol
Moles of N2 = mass / molar mass = 21 g / 28.02 g/mol = 0.749 mol
Next, we need to compare the stoichiometry of Mg and N2 in the balanced equation:
3Mg + N2 --> Mg3N2
According to the balanced equation, the ratio of Mg to N2 is 3:1. This means that for every 3 moles of Mg, we need 1 mole of N2 to fully react.
By comparing the moles available for each reactant, we can determine the limiting reagent. In this case, we have 1.186 moles of Mg and 0.749 moles of N2.
To find the limiting reagent, we divide the moles of each reactant by their stoichiometric coefficients:
Mg: 1.186 mol / 3 = 0.395 mol
N2: 0.749 mol / 1 = 0.749 mol
From the calculation, we can see that N2 has a lower value after dividing by the stoichiometric coefficient. Therefore, N2 is the limiting reagent.
The limiting reagent is the reactant that restricts the amount of product that can be formed. In this case, since N2 is the limiting reagent, it means that all of the N2 will be consumed, and Mg will be left over.
I hope this explanation helps! Let me know if you have any further questions.