An ice freezer behind a restaurant has a freon leak, releasing 47.97 g of C2H2F3Cl into the air every week. If the leak is not fixed, how many kilograms of fluorine will be released into the air over 6 months

0.554 kg

Explanation:

We want to find the amount of kilograms of fluorine that will be released into the air over 6 months.

Let's convert to weeks to get;

6 × 4 = 24 weeks

Let's find Mass leak rate of fluorine from the formula;

Mass leak rate = (fluorine mass in freon/molar mass of freon) × leak rate

Molar mass of freon = ((12 × 2) + (1 × 2) + (19 × 3) + (35.5)) = 118.5 g/mol

Thus;

Mass leak rate = ((19 × 3)/(118.5)) × 47.97 = 23.074 g/week

Total fluorine leaked in 6 months = 24 × 23.074 = 553.776 g = 0.554 kg