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Question: For the reaction, \(A + 2B \to C\), \(5\) moles of A and \(8\) moles of B will produce- A.\(5\) Mo...

For the reaction, A+2BCA + 2B \to C, 55 moles of A and 88 moles of B will produce-
A.55 Moles of C
B.44 Moles of C
C.88 Moles of C
D.1313 Moles of C

Explanation

Solution

We will solve this question by finding the limiting reagent in the reaction. The limiting reagent governs the amount of product produced as the product will be produced only until the limiting reagent is present in the reaction. The reactant which is present in less quantity than required to produce the product will be the limiting reagent. Then we will find the number of moles of C based on the number of moles of limiting reagent.

Complete step by step answer:
Given reaction is-A+2BCA + 2B \to C
According to the reaction, one mole of A reacts with 22 moles of B.
We can write it as-
The number of moles of B that react with one mole of A=22
But given that there are 55 moles of A so we can write-
The number of moles of B that react with 55 moles of A =2×52 \times 5
On solving, we get-
The number of moles of B that react with 55 moles of A =1010
But in the question it is given that there are 88 moles of B present in the reaction.
So B will limit the production of the product C as it is less than the required number.
This means that B is the limiting reagent. So the moles of C will depend on the moles of B present in the reaction.
According to the reaction, 22 moles of B produce 11 mole of C.
Then one mole of B will produce=12\dfrac{1}{2} moles of C
Since there are88moles of B then, 88 moles of B will produce=82\dfrac{8}{2} moles of C.
On solving, we get-
\Rightarrow 88 Moles of B will produce=44 moles of C

The correct answer is option B.

Note:
We can find the limiting agent using the formula-
\Rightarrow Reactant=Given molescoefficient of the reactant\dfrac{{{\text{Given moles}}}}{{{\text{coefficient of the reactant}}}} This will tell us which reactant has minimum value.
So on putting the values in the formula, we get-
\Rightarrow A=51\dfrac{5}{1} =55
\Rightarrow B=82=4\dfrac{8}{2} = 4
Here A>BA > B so B is the limiting reagent.