## Chemistry: Molecular Approach (4th Edition)

$$1.8 \times 10^{2} \space g$$
$CaCl_2$ : ( 35.45 $\times$ 2 )+ ( 40.08 $\times$ 1 )= 110.98 g/mol - Use all the information as conversion factors: $$5.5 \space L \times \frac{ 0.300 \space mol}{1 \space L} \times \frac{ 110.98 \space g}{1 \space mol} = 1.8 \times 10^{2} \space g$$