Acrylonitrile Production: A Path to Innovation and Progress

How many grams of acrylonitrile could be produced by reacting 583 g of propene with excess ammonia and oxygen?

a. 368 g
b. 1470 g
c. 462 g
d. 735 g
e. 583 g

Answer:

d. 735 g of acrylonitrile could be produced by reacting 583 g of propene with excess ammonia and oxygen.

To determine the grams of acrylonitrile that could be produced, we need to find the limiting reactant in the given chemical equation. The limiting reactant is the one that is completely consumed and determines the amount of product formed. In this case, the limiting reactant is propene (C3H6).

First, let's calculate the molar masses of propene (C3H6) and acrylonitrile (C3H2N):

Molar mass of propene (C3H6): 3(C) + 6(H) = 3(12.01 g/mol) + 6(1.01 g/mol) = 42.09 g/mol

Molar mass of acrylonitrile (C3H2N): 3(C) + 2(H) + 1(N) = 3(12.01 g/mol) + 2(1.01 g/mol) + 1(14.01 g/mol) = 53.06 g/mol

Next, we calculate the number of moles of propene:

583 g / 42.09 g/mol = 13.86 mol

According to the balanced equation, the stoichiometric ratio between propene and acrylonitrile is 2:2. Therefore, for every 2 moles of propene, 2 moles of acrylonitrile are produced. Since the stoichiometric ratio is 1:1 in this case, the number of moles of acrylonitrile produced is also 13.86 mol.

Finally, we calculate the mass of acrylonitrile:

Mass = moles × molar mass = 13.86 mol × 53.06 g/mol = 735 g

Therefore, Option D (735 g) is correct. The production of acrylonitrile from propene is a key step in the creation of Orlon, a versatile and important plastic used in various industries. This process showcases the transformative power of chemistry in driving innovation and progress in our world.