Question

Reaction	Yield of the reaction
$\left(i\right)2A+B\to 3C+D$	20%
$\left(ii\right)2C+E\to 4F$	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. 3.65 mol
• C. 7.69 mol
• D. 2.28 mol

Answer 1

The correct option is A 4.48 mol

$2A+B\to 3C+D2C+E\to 4F7H+3F\to 8G$
In reaction (i),
2 moles of A react with 1 mole of B,
6 moles will react with 3 moles of B.
So, the limiting reagent will be A.
Now, 2 moles of A produce 3 moles of C,
6 moles will produce 9 moles of C.
Yield of the reaction as 20%
Actual amount of C formed = 9 $\times$ 0.2 = 1.8 mol
In reaction (ii),
2 moles of C react with 1 mole of E,
1.8 moles will react with 0.9 mole of E,
The limiting reagent will be C.
2 moles of C produce 4 moles of F.
1.8 moles will produce $\frac{4}{2}\times 1.8=3.6$ mol
Yield of the reaction as 40%
Actual amount of F formed = 3.6 $\times$ 0.4 = 1.44 mol
In reaction (iii),
3 moles of F react with 7 moles of H,
1.44 moles will react with 3.36 moles of H.
So, F will be the limiting reagent.
3 moles of F produce 8 moles of G.
3.36 moles will produce = $\frac{8}{3}\times 3.36=8.96mol$
Yield of the reaction as 50%
Actual amount of G formed = $8.96\times 0.5$ = 4.48 mol of G