anonymous
  • anonymous
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
Chemistry
schrodinger
  • schrodinger
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anonymous
  • anonymous
Can you help with this one @Photon336
anonymous
  • anonymous
anonymous
  • anonymous

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Rushwr
  • 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 \[ mass = moles* molar mass\] \[mass = 2* 101 = 202g\]
anonymous
  • anonymous
thank you !

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