Question

In the given scheme of reactions, 10 moles of A is taken, while reagent B, E and H are taken in excess. Find the moles of G formed.
$\begin{array}{cc}\text{Reaction}& \text{Yield of reaction}\\ \text{(i)}2A+B\to 3C+D& 20\%\\ \text{(ii)}2C+E\to 4F& 40\%\\ \text{(iii)}7H+3F\to 8G& 50\%\end{array}$

• A. 1.4 mol
• B. 3.2 mol
• C. 2.8 mol
• D. 8.4 mol

Answer 1

The correct option is B 3.2 mol

$\begin{array}{cc}\text{Reaction}& \text{Yield}\\ \text{(i)}2A+B\to 3C+D& 20\%\\ \text{(ii)}2C+E\to 4F& 40\%\\ \text{(iii)}7H+3F\to 8G& 50\%\end{array}$
Initially, the number of mol of A present = 10
In reaction (i),
2 mol of A reacts to produce 3 mol of C
10 mol of A will produce = $\frac{3}{2}\times 10=15$ mol of C
Yield of reaction = 20 % = $\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100\%$
Actual yield obtained = $\frac{20}{100}\times 15$ = 3 mol of C
In reaction (ii),
2 mol of C produces 4 mol of F
3 mol of C will produce = $\frac{4}{2}\times 3$ = 6 moles
Actual yield obtained = $\frac{40}{100}\times 6$ = 2.4 mol of F
In reaction (iii),
3 mol of F reacts to produce 8 mol of G
2.4 mol of F will produce = $\frac{8}{3}\times 2.4$ = 6.4 mol of G
Actual yield obtained = $\frac{50}{100}\times 6.4$ = 3.2 mol of G