Ammonia, nh 3 , may react with oxygen to form nitrogen gas and water. 4 nh 3 ( aq ) + 3 o 2 ( g ) ⟶ 2 n 2 ( g ) + 6 h 2 o ( l ) if 3.55 g nh 3 reacts with 5.33 g o 2 and produces 0.450 l of n 2 , at 295 k and 1.00 atm, how many moles of n 2 can be produced by each reactant? nh 3 : mol n 2 o 2 : mol n 2 based on the theoretical yields, which reactant is limiting?

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

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

where are the choice options?

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.