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The reactant which is entirely consumed in reaction is known as limiting reagent. In the reaction 2A + 4B → 3C + 4D, when 5 moles of A react with 6 moles of B, then

(i) which is the limiting reagent?

(ii) calculate the amount of C formed?

Answers (1)

2A + 4B \rightarrow 3C + 4D

A and B are in the ratio- 2:4,  2 mols of 'A' require 4 mols of 'B' .

(i) For 5  mols of A, 10 mols of 'B' will be required using the ratio of 2:4.

i.e. 5 mol of A\times \frac{4\; mol\; of\; B}{2\; mol\; of\; A}=10 mol of

But only 6 mols of B is available therefore it is the limiting reagent

(ii), Amount of 'C' formed is depends on 'B' ,

4 mols of 'B' give 3 mols of C, the ratio is 4:3.

 Therefore, 6 mols of B will give, 6 mol of  B\times \frac{3\; mol\; of\; C}{4\; mol\; of\; C}=4.5\; moles\; of\; C

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