Question

In: Chemistry

The production capacity for acrylonitrile (C3H3N) in the United States exceeds 2 million pounds per year....

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) → 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?

Solutions

Expert Solution

Solution :-

Balanced reaction equation

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

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?

Solution :-

Lets first calculate the mass of the C3H3N produced from each reactant

Calculating from the propylene

(1.16 kg *1000 g/1 kg )*(1 mol/42.0804 g)*( 2mol C3H3N /2 mol C3H6)*(53.0637 g /1 mol ) = 1463 g

Calculating from NH3

(1.65 kg *1000g/1kg)*(1mol/17.03g)*(2 mol C3H3N/2mol NH3)* (53.0637 g /1 mol ) = 5141 g C3H3N

Calculating from O2

(1.78 kg *1000 g / 1kg)*(1mol / 32 g)*( 2 mol C3H3N/3mol O2)* (53.0637 g /1 mol ) = 2952 g C3H3N

C3H6 gives the lowest mass of the product

Therefore the C3H6 is the limiting reactant

Therefore the theoretical yield of the C3H3N = 1463 g

1463 g * 1 kg / 1000 g = 1.46 kg C3H3N

So it can produce 1.46 kg C3H6N

Part B

b) Calculating the mass of water produced

We know C3H6 is limiting reactat soo using the mole ratio we can calculate its mass

(1.16 kg *1000 g/1 kg )*(1 mol/42.0804 g)*( 6 mol H2O/2 mol C3H6)*(18.0148 g /1 mol ) = 1490 g

So the mass of the H2O that can be produced is 1490 g

1490 g * 1 kg / 1000 g = 1.49 kg H2O produced

c) Now lets calculate the mass of the O2 needed for the reaction

(1.16 kg *1000 g/1 kg )*(1 mol/42.0804 g)*( 3 mol O2 /2 mol C3H6)*(32 g /1 mol ) = 1323 g

So mass of O2 needed for reaction is 1323 g = 1.23 kg

So mass of O2 that can remain after reaction = 1.78 kg -1.23 kg = 0.55 kg O2 remain


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