JEE MAIN - Chemistry (2021 - 17th March Evening Shift - No. 15)
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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 : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u)
Answer
77
Explanation
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Stoichiometric moles of amide = 10$$-$$3 mol
Actual weight of amide = 10-3 $$ \times $$ 273 = 0.273 g
% yield = $${{0.210} \over {0.273}} \times 100$$
= 76.9%
$$ \simeq $$ 77%
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