Question

20. Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g) à 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.

Answer 1

First thing first: write and balance the euqation. In this case, it is already balance. Next, given the moles of the reactants, you should be able to identify which one of them is in excess and which one is limiting agent. The ratio of Mg to N2 is 3:1, meaning if you have 2 mol of N2, you should have had 6 mol of Mg. Hence mg is in excess (containing 8 instead of 6)..
Thirdly, consider the molar ratio of the entire reaction, basing from the limiting reatant (N2). From the equation N2 to Mg3N2 are in the ratio of 1 as to 1. So The product also has 2 mol.
Fourthly, use the formula mass=mol*molecular mass. Use your period table to find the total mass of the product and multiply it with the 0.2
Got it?

Answer 2

Good work, @Hoslos, but why to Multiply 0.2 in the end?

Answer 3

Oh , my mistake. Instead of 0.2 it should have been 2. Thanks @Gebooors.