Question

Consider the following balanced chemical reaction.

3A+4B+5C=2D

If 7 moles of A, 9 moles of B and 8 moles of C are given initially, what would be the mole fraction of D present after completion of reaction?

Answer 1

Answer: .4 is the right answer

Step-by-step explanation:

Answer 2

0.4mol is the answer.

explanation:

• Frst finding out limiting reagent.

• Here 8 moles of c  is the limting reagent because, if double amount that is 6mol A and 8mol B.

• We must take 10mol C but here, 7 mole A, 9moleB, take only 8mol c i.e, C is the limiting reagent.

 (i)

      $5 mol ------------- > 2\\8mol------------- > ?$

                     ∴ $D=3.2mol$

(ii)

                 $4molB--------- > 5molCrequire\\ ? < -----------8molC$

                     $6.4mol$ $used$ $left = 2.6mol$

(iii)

                    $3molA------- > 5molC$

                  $? < ----------8molC$

                    $4.9 mol$ $used$ $left=2.2mol$

• so total no:moles after raection= 8mol.

• mole fraction$=$ $\frac{no:moles of 'D'}{total No:mols}$

                             $=\frac{3.2}{8}$

                             $= 0.4$