Limiting Reagent

Question

Arcylonitrile, C₃H₃N, is the starting material for the production of a kind of synthetic fibre. It can be made from propylene, C₃H₆, by reaction with nitric oxide, NO, as follows:

4C₃H₆ + 6NO ⮕ 4C₃H₃N + 6H₂O + N₂

What mass of C₃H₃N can be made when 21.6 g of C₃H₆ react with 21.6 g of nitric oxide?

Answer 1

I'll do this one only because it has some stranger chemicals in it. But it is EXACTLY the same as the previous 4 or 5 problems that you've submitted. Again, you should be able to do these by yourself. So, follow along..

4C₃H₆ + 6NO ==> 4C₃H₃N + 6H₂O + N₂ ... balanced equation

mols C₃H₆ = 21.6 g C₃H₆ x 1 mol C₃H₆ / 42.08 g = 0.513 mols C₃H₆ (÷4->0.128)

mols NO = 21.6 g NO x 1 mol NO / 30.0 g = 0.720 mols NO (÷6->0.120)

These values (0.128 and 0.120) are very close to each other, but according the this calculation, NO would be limiting and as such will determine how much product can be made.

Theoretical yield of C₃H₃N

0.720 mols NO x 4 mols C₃H₃N / 6 mols NO = 0.480 mols C₃H₃N formed

0.480 mols C₃H₃N x 53.1 g C₃H₃N / mol C₃H₃N = 25.5 g C₃H₃N