## anonymous one year ago 20. The following is a Limiting Reactant problem: Magnesium nitride is formed in the reaction of magnesium metal with nitrogen gas in this reaction: 3 Mg(s) + N2(g)  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

1. anonymous

Can you help with this one @Photon336

2. anonymous

or @aaronq

3. anonymous

@Missiey

4. Rushwr

This the only equation we are gonna use here okai ? no. of moles = mass/ molar mass $n = \frac{ m }{ M }$ As we can see the molar ratios of Mg:N is 3:1 That means 3 moles of magnesium will react with 1 mole of nitrogen. When u have 2 moles of nitrogen that means it will react with 6 (2*3) moles of magnesium. But we have 8 moles of magnesium. So that means we have 2 moles (8-6) moles in excess. Now look at the stoichiometric number of magnesium nitride. It is 1 . Look at the molar ratio between nitrogen and magnesium nitride that is 1:1 That means one mole of nitrogen reacts to form 1 mole of magnesium nitride. Therefore no. of moles of nitrogen = no. of moles of magnesium nitride. Molar mass of magnesium nitride = 101gmol^-1 the no. of moles = 2moles $mass = moles* molar mass$ $mass = 2* 101 = 202g$

5. anonymous

thank you !