Question

In the following set of reactions:
$3E+4F\to 2G+3HG+3I\to 4J+7K4J+3L\to 9N+2O$
Starting with 12 mol of E, 8 mol of F, 5 mol of I and 4.66 mol of L. Find the moles of N formed. Assume yields of all reactions to be 100 %.

• A. 15.75 mol
• B. 13.98 mol
• C. 10.80 mol
• D. 25.60 mol

Answer 1

The correct option is B 13.98 mol

$3E+4F\to 2G+3HG+3I\to 4J+7K4J+3L\to 9N+2O$
In reaction (i),
3 moles of E react with 4 moles of F
12 moles will react with $\frac{4}{3}\times 12=16$ moles of F
So, the limiting reagent will be F.
Now, 4 moles of F produce 2 moles of G.
8 moles of F will produce $\frac{2}{4}\times 8=4$ moles of G.
In reaction (ii),
1 mole of G reacts with 3 moles of I.
4 moles will react with 12 moles of I.
Limiting reagent will be I.
3 moles of I produce 4 moles of J.
5 moles will produce $\frac{4}{3}\times 5=6.67$ moles of J.
In reaction (iii),
4 moles of J react with 3 moles of L.
6.67 moles will react with = 5 moles of L.
So, L will be the limiting reagent.
3 moles of L produce 9 moles of N.
4.66 moles will produce
$=\frac{9}{3}\times 4.66=13.98mol$