Respuesta :
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
Answer:
The limiting reactant is NH3
Theoretical there can be produced 0.104 moles of N2
The actual yield is 0.0186 moles N2
Explanation:
Step 1: Data given
Mass of NH3 = 3.55 grams
Mass of O2 = 5.33 grams
Volume of N2 = 0.450 L
Temperature = 295 K
Pressure = 1.00 atm
Molar mass NH3 = 17.03 g/mol
Molar mass O2 = 32 g/mol
Step 2: The balanced equation
4 NH3(aq) + 3O2(g) ⟶ 2N2(g) + 6H2O(l)
Step 3: Calculate moles of NH3
Moles NH3 = Mass NH3 / molar mass NH3
Moles NH3 = 3.55 grams / 17.03 g/mol
Moles NH3 = 0.208 moles
Step 4: Calculate moles of O2
Moles O2 = 5.33 grams / 32.0 g/mol
Moles O2 = 0.167 moles
Step 5: Calculate the limiting reactant
For 4 moles NH3 , we need 3 moles O2 to produce 2 mol of N2 and 6 moles of H2O
NH3 is the limiting reactant. It will completely be consumed (0.208 moles).
O2 is in excess. There will be consumed 3/4 * 0.208 = 0.156 moles O2
There will remain 0.167 - 0.156 = 0.011 moles
Step 6: Calculate moles of N2
For 4 moles NH3 , we need 3 moles O2 to produce 2 mol of N2 and 6 moles of H2O
For 0.208 moles NH3 we produce 0.208/2 = 0.104 moles of N2 ( = the theoretical yield)
Step 7: Calculate volume of N2
p*V = n*R*T
⇒ p = the pressure =1.00 atm
⇒ V = the volume of N2 = TO BE DETERMINED
⇒ n= the number of moles of N2 = 0.104 moles
⇒ R = the gas constant = 0.08206 L*atm/K*mol
⇒ T = the temperature = 295 K
V = (n*R*T)/p
V = (0.104*0.08206*295)/1
V = 2.52 L =( theoretical yield)
Step 8: Calculate the % yield
% yield = actual yield / theoretical yield
% yield = (0.45 L / 2.52 L) *100%
% yield = 17.9 %
Theoretical there can be produced 0.104 moles of N2
The actual yield is 0.0186 moles N2
