Question

Reaction	Yield of the reaction
$\left(i\right)2A+3B\to 3C+D$	80%
$\left(ii\right)2C+E\to 4F$	40 %
$\left(iii\right)3F+7H\to 8G$	50%

Starting with 6 mol of A, 9 mol of B, 4 mol of E and 15 mol of H, the number of moles of G formed will be

• A. 4.48 mol
• B. 7.68 mol
• C. 5.28 mol
• D. 6.68 mol

Answer 1

The correct option is B 7.68 mol

Given,
(i) $2A+3B\to 3C+D$
(ii) $2C+E\to 4F$
(iii) $7H+3F\to 8G$

In reaction (i),
2 moles of A react with 3 moles of B, 6 moles of A will react with 9 moles of B.
Now, 2 moles of A produce 3 moles of C. Hence, 6 moles of A should produce 9 moles of C.
But, the yield of the reaction is given as 80%
Thus, the actual amount of C formed $=9\times 0.8=7.2mol$

In reaction (ii),
2 moles of C react with 1 mole of E, 7.2 moles of C will react with 3.6 moles of E.
We are given 4 moles of E.
So, the limiting reagent is C here.
2 moles of C produce 4 moles of F.
7.2 moles of C will produce $\frac{4}{2}\times 7.2=14.4$ mol of F
Here, the yield of the reaction is given as 40%
Actual amount of F formed = 14.4 $\times$ 0.4 = 5.76 mol

In reaction (iii),
3 moles of F react with 7 moles of H, 5.76 moles of F will react with 13.44 moles of H.
So, F is the limiting reagent here.
3 moles of F produce 8 moles of G.
5.76 moles will produce $=\frac{8}{3}\times 5.76=15.36mol$
Again, yield of the reaction is given as 50%
Actual amount of G formed = $15.36\times 0.5$ = 7.68 mol of G