Based upon the previous question, what is the THEORETICAL YIELD (in grams) of Mg3N2 if you start with 10.0 grams of each reactant?
previous question:consider the magnesium nitride addition reaction again. This time, suppose you start with 10.0 g of each reactant. What is the maximum number of MOLES of Mg3N2 that could be produced in this case? Remember that the reactant that runs out first is called the LIMITING REACTANT.
To calculate the theoretical yield of Mg3N2 in grams, we first need to determine the limiting reactant based on the information provided in the previous question.
The limiting reactant is the reactant that will be completely consumed first and, therefore, determines the maximum amount of product that can be formed.
Since we know that the reaction produces 1 mole of Mg3N2 for every 3 moles of Mg used and 1 mole of Mg3N2 for every 1 mole of N2 used, we can calculate the number of moles of Mg and N2 available by using their respective molar masses.
The molar mass of Mg is 24.31 g/mol, and the molar mass of N2 is 28.02 g/mol.
For Mg:
10.0 g of Mg / (24.31 g/mol) = 0.411 mol of Mg
For N2:
10.0 g of N2 / (28.02 g/mol) = 0.357 mol of N2
To determine the limiting reactant, we compare the ratio of moles of each reactant to their stoichiometric coefficient in the balanced equation. In this case, the ratio is 3:1 (3 moles of Mg to 1 mole of N2).
Considering this ratio, we have 0.411 mol of Mg and 0.357 mol of N2. Since the ratio of moles of Mg to N2 is greater than 3:1, N2 is the limiting reactant.
Now we can calculate the maximum number of moles of Mg3N2 that can be produced using the limiting reactant:
0.357 mol of N2 * (1 mol of Mg3N2 / 1 mol of N2) = 0.357 mol of Mg3N2
Finally, we need to calculate the theoretical yield of Mg3N2 in grams by multiplying the number of moles of Mg3N2 by its molar mass:
0.357 mol of Mg3N2 * (100.95 g/mol) = 36.15 g of Mg3N2
Therefore, the theoretical yield of Mg3N2 is 36.15 grams when starting with 10.0 grams of each reactant.