Answer
The amounts of $A=0.90 \space mol$ and of $B=1.20 \space mol$.
Work Step by Step
Let us consider a reaction $A \to 2B$
Case I: To find $A$.
We are given that the data as follows: Initial moles; $n_{A0}=1.50 mol$ and Extent for a reaction $\triangle \xi = 0.60 mol$
Now, $n_{A}=n_{A0} -v_{A} \triangle \xi$
Plug in the data .
$n_{A}=1.50 mol -0.60 mol=0.90 \space mol$
Case 2: To find $B$.
We are given that the data as follows: Initial moles; $n_{B0}=0 \space mol$ and Extent for a reaction $\triangle \xi = 0.60 mol$
Now, $n_{B}=n_{B0} -v_{B} \triangle \xi$
Plug in the data .
$n_{B}=0 mol +2(0.60 mol)=1.20 \space mol$
Therefore, the amounts of $A=0.90 \space mol$ and of $B=1.20 \space mol$.
