Answer
0.154 mole of $Cl^-$ and 0.154 mole of $Na^+$
Work Step by Step
1. Identify the conversion factors:
$Na^+$ has a charge of +1: 1 mole $Na^+$ = 1 Eq $Na^+$
$Cl^-$ has a charge of -1: 1 mole $Cl^-$ = 1 Eq $Cl^-$
$154 \space mEq \space Na^+= 1 \space L \space of \space solution$
$154 \space mEq \space Cl^-= 1 \space L \space of \space solution$
$1000 \space mEq = 1 \space Eq$
2. Calculate the amount of moles.
$$1.00 \space L \space of \space solution \times \frac{154 \space mEq \space Na^+}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space Na^+}{1 \space Eq \space Na^+} = 0.154 \space mole \space Na^+$$
$$1.00 \space L \space of \space solution \times \frac{154 \space mEq \space Cl^-}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space Cl^-}{1 \space Eq \space Cl^-} = 0.154 \space mole \space Cl^-$$
