Answer
Freon-12: $ 1180 \ g$
Tetrafluoroethane: $ 848 \ g$
Work Step by Step
1. Calculate the heat necessary to reduce the temperature to 0.
$$16 oz \times \frac{28 \ g}{1 \ oz} \times \frac{1 \ mol}{18.0 \ g}= 24.9 \ mol$$
q = ( 24.9 mol )( 75.3 J/(mol K))( 18 K )
q = $ 3.37 \times 10^{4} $ J
2. Now, find the energy to freeze the water.
$q = n \Delta H_{fus} = 24.9 \ mol \times 6010 \ J/mol = 1.496 \times 10^5 \ J$
Total energy required: $3.37 \times 10^4 + 1.496 \times 10^5 = 1.83 \times 10^5 \ J$
3. Find the amount of mass required of each compound:
Freon-12: $$\frac{1.83 \times 10^5 \ J}{155 \ J/g} = 1180 \ g$$
Tetrafluoroethane $$\frac{1.83 \times 10^5 \ J}{215.9 \ J/g} = 848 \ g$$
