The following is a Limiting Reactant problem:
Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g) Mg3N2(s)
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.
I need help with this
There are shorter ways to do this but they are harder to explain. I do them the long way this way.
mols N2 = grams/molar mass = ?
mols Mg = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols N2 to mols Mg3N2.
Do the same for mols Mg to mols Mg3N2.
Probably those two values will not agree which means one is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent responsible for that value is the LR.
To find grams, take the smaller number of the two and convert to grams. grams = mols x molar mass = ?
By the way, that is the theoretical yield of the reaction.
ok thank you
To solve this limiting reactant problem, we need to determine which reactant will be completely consumed and which reactant will be present in excess. The reactant that is completely consumed is called the limiting reactant, and it determines the maximum amount of product that can be formed.
Step 1: Write and balance the equation:
3 Mg(s) + N2(g) → Mg3N2(s)
Step 2: Convert moles to grams:
Given:
2.0 mol N2
8.0 mol Mg
Now, we need to convert moles of each reactant to grams using their respective molar masses.
Molar mass of N2: 2 * (14.01 g/mol) = 28.02 g/mol
Molar mass of Mg: 24.31 g/mol
From the given amount of N2:
Mass of N2 = 2.0 mol * 28.02 g/mol = 56.04 g
From the given amount of Mg:
Mass of Mg = 8.0 mol * 24.31 g/mol = 194.48 g
Step 3: Calculate the stoichiometric ratio:
Using the balanced equation, we can determine the stoichiometric ratio of Mg to N2. In this case, it is 3:1.
Step 4: Determine the limiting reactant:
Compare the moles of each reactant to their respective stoichiometric ratios. The reactant that has a smaller number of moles compared to its stoichiometric ratio will be the limiting reactant.
For N2:
Stoichiometric ratio: 1 mol N2 = 3 mol Mg
Number of moles available: 2.0 mol N2
For Mg:
Stoichiometric ratio: 3 mol Mg = 1 mol N2
Number of moles available: 8.0 mol Mg
From the above comparison, it is clear that N2 has fewer moles compared to its stoichiometric ratio, making it the limiting reactant.
Step 5: Calculate the moles of product:
Knowing that N2 is the limiting reactant, we can use the stoichiometric ratio from the balanced equation to determine the moles of product formed.
Stoichiometric ratio: 1 mol N2 = 1 mol Mg3N2
Therefore, the number of moles of Mg3N2 formed will be equal to the number of moles of N2:
Moles of Mg3N2 formed = Moles of limiting reactant (N2) = 2.0 mol
Step 6: Convert moles of product to grams:
Finally, we can convert moles of Mg3N2 formed to grams using its molar mass.
Molar mass of Mg3N2: 24.31 g/mol (Mg) + 14.01 g/mol (N) * 2 = 100.95 g/mol
Mass of Mg3N2 formed = 2.0 mol * 100.95 g/mol = 201.9 g
Therefore, 201.9 grams of product (Mg3N2) will be formed from 2.0 mol of N2 and 8.0 mol of Mg.