Question

For the given sequence of the reactions:
$A+2B\to 2C+D\text{yield}=50\%$
$C+12E\to 4F+3G\text{yield}=20\%$
$2F+3H\to 2I+J\text{yiled}=30\%$
If $10$ moles of $A$ reacts with $21$ moles of $B$ then find the number of moles of $I$ produced by $11$ mol of $H$ if $E$ is present in excess.

• A. $2.20$
• B. $7.33$
• C. $2.40$
• D. $8$

Answer 1

The correct option is A $2.20$

$A+2B\to 2C+D\text{yield}=50\%$
Ratio of number of moles to coefficients:
$\frac{n_A}{1}=10$
$\frac{n_B}{2}=10.5$
A is the limiting reagent in the first reaction.
$\frac{n_A\left(\text{consumed}\right)}{1}\times \text{yield}\%=\frac{n_C\left(\text{produced}\right)}{2}$
$\Rightarrow n_C=10\text{mol}$

$C+12E\to 4F+3G\text{yield}=20\%$
$\frac{n_C\left(\text{consumed}\right)}{1}\times \text{yield}\%=\frac{n_F\left(\text{produced}\right)}{4}$
$\Rightarrow n_F=8mol$
$2F+3H\to 2I+J\text{yield}=30\%$
Ratio of number of moles to coefficients:
$\frac{n_F}{2}=4$
$\frac{n_H}{3}=\frac{11}{3}$
$H$ is limiting reagent.So
$\frac{n_H\left(\text{consumed}\right)}{3}\times \text{yield}\%=\frac{n_I\left(\text{produced}\right)}{2}$
$\Rightarrow n_I=2.2\text{mol}$