Question

6 moles of A and 20 moles of B are taken after the reaction. Calculate limiting reagent ?and Moles of AB and A2B3 ?
explain step by step
pls​

Answer 1

Answer:

Hence, moles of AB is 1 (Ans)

Explanation:

Here 6 moles of A and 20 moles of B are taken part of reaction.

It's clear that 1 mole of A can react with only one mole of B.

So, here 6 moles of A will react only with 6 moles of B.

Remaining (20-6)= 14 moles of B will remain unreactive.

Hence,

here limiting reagent is A, and

excess reagent is B.

2nd part,  

Calculate moles of AB and A2B3

In case of A2B3

First,

we have to balance the equation

2A+ 3B = A2B3

no. Of moles : 2A  +  3 B =   A2B3

                          6   + 20  =  ?( required  value)

Stoichiometric : 2      3 = 1

Coefficient  

Now mole mole analysis is  

NA/2 = NB/3 = NA2B3/1

As here limiting reagent is A, so we compare NA/2 to NA2B3,  

So, NA/2 = NA2B3/1

    6/2 = NA3B3/1

 So, NA2B3 = 3

Hence, moles of A2B3 is 3

In case of AB

First, we have to balance the equation

A+ B = AB

no. Of moles : A  +   B =   AB

                          6 + 20  =  ?( required  value)

Stoichiometric : 6/1.  20/1  = 1

Coefficient  

Now mole mole analysis is  

NA/6= NB/20 = NAB/1

As here limiting reagent is A, so we compare NA/2 to NAB

So, NA/6= NAB/1

    6/6 = NAB/1

 So, NAB= 1

Hence, moles of AB is 1 (Ans)