Consider once again the magnesium nitride addition reaction from the previous question. This time, suppose you start with 10.0 g N2 and an unlimited supply of Mg. What is the maximum number of MOLES of Mg3N2 that could be produced in this case?
To determine the maximum number of moles of Mg3N2 that could be produced, we need to calculate the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed.
The balanced chemical equation for the reaction is:
3Mg + N2 â Mg3N2
From the equation, we can see that the stoichiometric ratio between Mg3N2 and N2 is 1:1. This means that one mole of Mg3N2 can be formed for every mole of N2 consumed.
First, we need to convert the given mass of N2 to moles. The molar mass of N2 is 28.02 g/mol.
Number of moles of N2 = Mass of N2 / Molar mass of N2
= 10.0 g N2 / 28.02 g/mol
â 0.357 mol N2
Since the stoichiometric ratio between Mg3N2 and N2 is 1:1, the maximum number of moles of Mg3N2 that can be produced is also approximately 0.357 moles.
Therefore, the maximum number of moles of Mg3N2 that could be produced in this case is approximately 0.357 moles.