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If mg + n2 → Mg3N2rnrn277 grams of mg are reacted with 0.859 moles of N2, how many moles of Mg3N2 will be produced?
Accepted Answer
The balanced chemical equation for the reaction is:
3Mg + N2 → Mg3N2
To determine the limiting reactant, we will calculate the moles of each reactant:
moles of Mg = 277 g / 24.31 g/mol = 11.39 mol
moles of N2 = 0.859 mol
The mole ratio of Mg to N2 is 3:1. Therefore, the limiting reactant is N2.
According to the stoichiometry of the balanced equation, 1 mole of N2 reacts with 3 moles of Mg to produce 1 mole of Mg3N2. Therefore, 0.859 moles of N2 will produce:
moles of Mg3N2 = 0.859 mol N2 × (1 mol Mg3N2 / 1 mol N2) = 0.859 mol Mg3N2
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