Question

The production capacity for acrylonitrile (C3H3N) in the United States exceeds 2 million pounds per year. Acrylonitrile, the building block for polyacrylonitrile fibers and a variety of plastics, is produced from gaseous propylene, ammonia, and oxygen. 2 C3H6(g) + 2 NH3(9) + 3 02(g) - 2 C3H3N(g) + 6 H20(g) (a) What mass of acrylonitrile can be produced from a mixture of 1.25 kg of propylene (C3H6), 1.66 kg of ammonia, and 2.00 kg of oxygen, assuming 100% yield? (b) What mass of water is produced? Which starting materials are left in excess? (Select all that apply.) ப ந What mass of propylene is left in excess? What mass of ammonia is left in excess? What mass of oxygen is left in excess? Noed Help?

Any help would be greatly appreciated, I've tried to solve it so many times and I'm running out of chances.

Answer 1

PART (a):

Given reaction:

2C3H6 + 2NH3 + 302 → 2C3H3N + 6H2O

We have, Molar mass of C₃H₆= 42.08 g/mol

Molar mass of ammonia (NH₃) = 17.031 g/mol

Molar mass of oxygen (O₂) = 32 g/mol

Therefore, No.of moles of propylene:

1.25 kg x (100) 42.08 g/mol n C3H6 =

1250 g nczHs 42.08 g/mol

ncha = 29.705 mol

No.of moles of ammonia:

n NH3- 1.66 kg x (1000 17 g/mol

1660 g N NH3 – 17 g/mol

nNH, = 97.647 mol

No.of moles of oxygen:

2.00 kg x (1002) noza 32 g/mol

2000 g noz 32 g/mol

no, = 62.5 mol

As we can see, Here propylene (C₃H₆) will be the limiting reagent and other reactants (oxygen and ammonia) are in excess.

Hence, all stoichiometric calculations should be based on limiting reagent, which is propylene.

• From balanced reaction we have, 2 moles of propylene produces 2 moles of acrylonitrile.
• Hence, 1 mole of propylene produces 1 moles of acrylonitrile.
• Therefore, 29.705 moles of propylene produces 29.705 moles of acrylonitrile.

Therefore, Mass of acrylonitrile​​​​​​​ produced is given by:

mcH N = 29.705 mol x (53.06 g/mol)

mc, H, N = 1576.2 g

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PART (b):

• From balanced reaction we have, 2 moles of propylene produces 6 moles of water.
• Hence, 1 mole of propylene produces 3 moles of water.
• Therefore, 29.705 moles of propylene produces 3 x 29.705 moles of water = 89.12 moles

Therefore, Mass of water produced is given by:

m H,0 = 89.12 mol x (18 g/mol

6 T09T = O'Hur

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PART (c):

• Ammonia and oxygen are in excess.

Refer part (a) for explanation.

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PART (d):

Propylene is the limiting reagent. Hence it will be completely consumed during the reaction.

So nothing is left after reaction.

So mass of Propylene ​​​​​​​left in excess = 0 g.

m = 0 g

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PART (e):

No.of moles of ammonia in the reaction mixture = 97.647 moles

No.of moles of ammonia reacted = 29.705 moles

No.of moles of ammonia left = 97.647 moles - 29.705 moles = 67.94 moles

Therefore, Mass of ammonia left in excess is given by:

mNH= (67.94 mol) x (17 g/mol)

- MNH, = 1155 g

>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>

PART (f):

No.of moles of oxygen in the reaction mixture = 62.5 moles

No.of moles of ammonia reacted = (3/2) x 29.705 moles = 44.558 moles

No.of moles of ammonia left = 62.5 - 44.558 = 17.94 moles

Therefore, Mass of oxygen left in excess is given by:

mo, = (17.94 mol) x (32 g/mol)

-. mo, = 574 g ​​​​​​​