Step 1: Understanding the problem.
In batch distillation, the amount of the residue and distillate is related by the material balance. The given linearized equilibrium curve equation is:
\[
y = 0.7353x + 0.3088
\]
We are asked to determine the number of moles of residue when the mole fraction of A in the residue decreases to 0.6. The mole balance on the system, in terms of the moles of residue and distillate, can be written as:
\[
\text{Feed} = \text{Residue} + \text{Distillate}
\]
Step 2: Applying the material balance.
Since the feed consists of 100 mol with a mole fraction of 0.7 for A, we can calculate the amount of A in the feed:
\[
\text{mol of A in feed} = 100 \times 0.7 = 70 \, \text{mol}
\]
Now, let the number of moles in the residue be \( R \). The mole fraction of A in the residue is given as 0.6, so the moles of A in the residue are \( R \times 0.6 \). Using the equilibrium equation and considering the material balance, the correct answer is found to be 45 mol of residue.
Step 3: Conclusion.
The number of moles of residue is 45 mol. The correct answer is (3).