Answer:
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
