Answer
0.060 mole of $K^-$ and 0.060 mole of $Cl^-$
Work Step by Step
1. Identify the conversion factors:
$K^+$ has a charge of +1: 1 mole $K^+$ = 1 Eq $K^+$
$Cl^-$ has a charge of -1: 1 mole $Cl^-$ = 1 Eq $Cl^-$
$40. \space mEq \space K^+= 1 \space L \space of \space solution$
$40. \space mEq \space Cl^-= 1 \space L \space of \space solution$
$1000 \space mEq = 1 \space Eq$
2. Calculate the amount of moles.
$$1.5 \space L \space of \space solution \times \frac{40. \space mEq \space K^+}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space K^+}{1 \space Eq \space K^+} = 0.060 \space mole \space K^+$$
$$1.5 \space L \space of \space solution \times \frac{40. \space mEq \space Cl^-}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space Cl^-}{1 \space Eq \space Cl^-} = 0.060 \space mole \space Cl^-$$
