Question

Consider the above reaction. The percentage yield of amide product is ......... (Round off to the Nearest Integer).

(Given : Atomic mass : C : 12.0 u, H : 1.0u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)

Answer 1

∴ Theoretical amount of given product formed

= \(\frac{273}{140.5}\) x 0.140  0.272 gm

But its actual amount formed is 0.210 gm. Hence, the percentage yield of product.

OR

by mol-mol analysis = 10^–3 mol.

Theoritical mass of product = 10^–3 × 273 = 273 × 10^–3 g

Observed mass of product = 210 × 10^–3 g

%yield of product = \(\frac{210\times10^{-3}}{273\times10^{-3}}\) x 100 = 76.9% = 77