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 (g) + 3 O2 (g) â 2 C3H3N (g) + 6 H2O (g)

(a) What mass of acrylonitrile can be produced from a mixture of 1.16 kg of propylene (C3H6), 1.65 kg of ammonia, and 1.78 kg of oxygen, assuming 100% yield?

(b) What mass of water is produced?

(c) What mass of oxygen is left in excess?

This question we will solve by calculations based on the stoichiometry of the balanced chemical equation which gives us all the information required to know the quantities produced and reacted based on their molar ratios.

First we will need the molecular weights of the reactants to calculate the number of moles of each reactant and determine if there is a limiting reagent, and from there we can learn about the moles and masses of the products.

2 C3H6 (g) + 2 NH3 (g) + 3 O2 (g) ⇒ 2 C3H3N (g) + 6 H2O (g)

MW C3H6 : 42.08 g/mo MW C3H3N : 53.06 g/mol

MW NH3 : 17.03 g/mol MW H2O : 18.02 g/mol

MW O2: 32 g/mol

Moles of reactants:

Convert the masses given to grams since we have the molar masses in grams. The number of moles, n, is calculated by dividing the mass into the molecular weight.

n C3H6 = (1.16 Kg x 1000 g / Kg) / 42.08 g/mol = 27.56 mol

n NH3 = (1.65 kg x 1000 g / / Kg) / 17.03 g/mol = 96.89 mol

n O2 = (1.78 Kg x 1000 g / Kg) / 32 g/mol = 55.63 mol

from the stoichiometry of the reaction we know propylene and ammonia react 2: 2 so propylene is the limiting reagent:

(2 mol NH3 / 2 mol C3H6) x 27.56 mol C3H6 = 27.56 mol NH3 (required to react with the 27.56 mol C3H6 and we have plenty (96.98 mol)

The stoichiometry of the reaction also confirms that O2 is in excess:

(3 mol O2 / 2 mol C3H6) x 27.56 mol C3H6 = 41.34 mol O2 (required to react completely with 27.56 mol C3H6).

(a) Again from the balanced chemical reaction we know the mol proportions reactants to product, thus mol C3H3N (1: 1) produced:

(2 mol C3H3N / 2 mol C3H6) x 27.56 mol C3H6 = 27.56 mol C3H3N

The mass of acrylonitrile will be given by multiplying the molecular weight of the mol produced assuming a 100 % yield:

55.12 g/mol x 27.56 mol = 1,519 g = 1.51 Kg

(b) The calculation to obtain the mass of water will be performed in a similar manner:

(6 mol H2O / 2 mol C3H6) x 27.56 mol C3H6 = 82.68 mol H2O produced

82.68 mol x 18 g/mol = 1,488 grams = 1.49 Kg

(c) The mass of O2 left will be obtained from the number of moles in excess:

mol O2 originally present = 1.78 x 1000 g/Kg / 32 g/mol = 55.62 mol

mol O2 in excess = mol O2 initially - mol reacted

from above we know 41.34 mol are required to react with our limiting reagent, C3H6:

mol O2 in excess = 55.62 mol - 41.34 mol = 14.29 mol

mass oxygen in excess = 32 g/mol x 14.29 mol = 457.12 g = 0.457 Kg