Question

9 mol of D and 14 mol of E are allowed to react in a closed vessel according to the given reactions. Calculate the number of mol of B formed in the end of reaction, if 4 mol of G are present in reaction vessel.
(Percentage yield of reaction is mentioned in the reaction)
$3D+4E\stackrel{80\%}{\to }5C+A$
$3C+5G\stackrel{50\%}{\to }6B+F$

• A. 2.4 mol
• B. 6.7 mol
• C. 4.8 mol
• D. 1.2 mol

Answer 1

The correct option is A 2.4 mol

$\underset{9}{3D}+\underset{14}{4E}\stackrel{80\%}{\to }5C+A$
3 mol of D reacts to give 5 mol of C which gives a yield of 80%.

For D, $\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{9}{3}=3$

For E, $\frac{\text{Initial number of moles}}{\text{stoichiometric coefficient}}=\frac{14}{4}=3.5$

Thus, D is the limiting reagent,
Number of moles of C produced by 9 moles of D $=\frac{5}{3}\times 9\times 0.8=12mol$

Similarly,
$\underset{12}{3C}+\underset{4}{5G}\stackrel{50\%}{\to }6B+F$
Here, G is the limiting reagent,
Number of moles of B produced by 4 moles of G $=\frac{6}{5}\times 4\times 0.5=2.4mol$