21 g each of Li (molar mass = 7 g/mol) and N2 (molar mass = 28 g/mol) are placed in a reaction vessel. What is the maximum mass of Li3N (molar mass = 35) that can be produced?

Question

Bubblegum

Chemistry

16 December, 16:50

21 g each of Li (molar mass = 7 g/mol) and N2 (molar mass = 28 g/mol) are placed in a reaction vessel. What is the maximum mass of Li3N (molar mass = 35) that can be produced?

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Blaise · Answer 1

Lithium reacts with nitrogen based on the following equation:

6Li + N2 ... > 2Li3N

This means that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N. In other words, 42 grams of Li are required to react with 28 grams of N2 to produce 70 grams of Li3N

Therefore, 21 grams of Li need to react with 14 grams of N2.

Based on this, lithium is our limiting reagent for this reaction and we will base the calculations on the amount of lithium.

amount of Li3N = (21x70) / 42 = 35 grams