AriellaBrown7
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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)?

Respuesta :

JoshEast
[tex]3 Mg_{(s)} + N _{2} _{(g)} ----\ \textgreater \ Mg_{3} N _{2} _{(s)} [/tex]

mole ratio of   Mg : Mg₃N₂   =  3 b:  1

∴  mole of  Mg₃N₂ =  [tex] \frac{mole of magnesium}{3} [/tex]
                  
                              =  [tex] \frac{8.0 mol}{3} [/tex]
                        
                              =  2.67 mol

mass of Mg₃N₂  =  mole  *  molar mass
                          =  2.67 mol  *  ((3 * 24) + (2 * 14)) g / mol
                          =  266.67 g


Dejanras

Answer is: 201.86 grams of product are formed.

Balanced chemical rection: 3Mg(s) + N₂(g) → Mg₃N₂(s).

n(N₂) = 2.0 mol; amount of nitrogen.

n(Mg) = 8.0 mol: amount og magnesium.

From balanced reaction: n(Mg) : n(N₂) = 3 : 1.

For 8 moles of magnesium:

8 mol : n(N₂) = 3 : 1.

n(N₂) = 2.66 mol.

There is not enough nitrogen to react completely with magnesiu, so nitrogen is limiting reactant.

From balanced reaction: n(N₂) : n(Mg₃N₂) = 1 : 1.

n(Mg₃N₂) = 2 mol.

m(Mg₃N₂) = n(Mg₃N₂) · M(Mg₃N₂).

m(Mg₃N₂) = 2 mol · 100.93 g/mol.

m(Mg₃N₂) = 201.86 g; mass of magnesium nitride.

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