Answer:
The limiting reactant is NH₃
0.0186moles of N₂ are the one produced by the limiting reactant
0.020 moles of N₂ are the one produced by the reactant in excess
Explanation:
This is the reaction
4NH₃ + 3O₂ → 2N₂ + 6H₂O
We should calculate the moles of each reactant
Mass / Molar mass = Moles
3.55 g / 17g/m = 0.208 moles NH₃
5.33 g / 32g/m = 0.166 moles O₂
4 moles of ammonia react with 3 moles of oxygen
0.208 moles of ammonia react with (0.208 .3)/4 = 0.156 moles O₂
We have 0.166 moles of O₂ and we need 0.156 moles, so O₂ is the reactant in excess.
3 moles of O₂ react with 4 moles of NH₃
0.166 moles of O₂ react with (0.166 . 4)/ 3 = 0.221 moles
We have 0.208 moles NH₃ and we need 0.221, so NH₃ is the limiting reactant.
To know the moles of N₂, let's apply the Ideal Gas Law
P.V =n.R.T
1atm . 0.450L = n . 0.082 . 295K
0.450 / (0.082 .295) = 0.0186 moles
If we have 100 % yield reaction:
4 moles NH₃ make 2 moles N₂
0.208 moles NH₃ make (0.208 .2)/4 = 0.104 moles
So the % yield reaction is.
0.104 moles ___ 100%
0.0186 moles ___ 17.9%
0.0186moles of N₂ are the one produced by the limiting reactant.
3 moles of O₂ produce 2 moles N₂
0.166 moles O₂ produce (0.166 .2)/3 = 0.111 moles
Now, we apply the yield.
100% ____ 0.111 moles
17.9% = 0.020 moles
