How many grams of product are formed from 2.0 mol of N2 (g) and 8.0 mol of Mg(s)? Show all calculations leading to an answer.
This is a limiting reagent problem. You know that when amounts are given for BOTH reactants.
3Mg + N2 ==> Mg3N2
a. Convert mols Mg to mols product.
b. Convert mols N2 to mols product.
These answers are likely to be different; if so then both can't be right. The correct value in limiting reagent problems is ALWAYS the smaller number and the reagent responsible for that is the limiting reagent.
Use the smaller number of mols for the product and convert that to grams product. That will be done with g = mols Mg3N2 x molar mass Mg3N2.
Post your work if you get stuck.
To determine the number of grams of product formed, we need to use stoichiometry and convert the given moles of reactants to moles of product, and then convert the moles of product to grams.
The balanced chemical equation for the reaction between N2 and Mg is:
N2(g) + 3Mg(s) → 2Mg3N2(s)
From the balanced equation, we can see that 1 mole of N2 reacts with 3 moles of Mg to produce 2 moles of Mg3N2. Therefore, the mole ratio between N2 and Mg3N2 is 1:2.
1) Convert the given 2.0 moles of N2 to moles of Mg3N2 using the mole ratio:
2.0 mol N2 × (2 mol Mg3N2 / 1 mol N2) = 4.0 moles of Mg3N2
2) Next, convert the given 8.0 moles of Mg to moles of Mg3N2 using the mole ratio:
8.0 mol Mg × (2 mol Mg3N2 / 3 mol Mg) = 5.33 moles of Mg3N2
Note: We converted from moles of Mg to moles of Mg3N2 using the mole ratio of 2 moles of Mg3N2 per 3 moles of Mg.
3) Finally, calculate the molar mass of Mg3N2, which is the sum of the atomic masses of Mg and N multiplied by their respective subscripts in the formula:
Molar mass of Mg3N2 = (3 × atomic mass of Mg) + (2 × atomic mass of N)
= (3 × 24.31 g/mol) + (2 × 14.01 g/mol)
= 72.93 g/mol + 28.02 g/mol
= 100.95 g/mol
4) Convert the moles of Mg3N2 to grams using the molar mass:
5.33 mol Mg3N2 × (100.95 g Mg3N2 / 1 mol Mg3N2) = 537.8 grams of Mg3N2
Therefore, approximately 537.8 grams of the product Mg3N2 would be formed from 2.0 moles of N2 and 8.0 moles of Mg.