## General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

0.154 mole of $Cl^-$ and 0.154 mole of $Na^+$
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. $$1.00 \space L \space of \space solution \times \frac{154 \space mEq \space Na^+}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space Na^+}{1 \space Eq \space Na^+} = 0.154 \space mole \space Na^+$$ $$1.00 \space L \space of \space solution \times \frac{154 \space mEq \space Cl^-}{1 \space L \space of \space solution} \times \frac{1 \space Eq}{1000 \space mEq} \times \frac{1 \space mole \space Cl^-}{1 \space Eq \space Cl^-} = 0.154 \space mole \space Cl^-$$