Question

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

Answer 1

The amount of fluorine released in 6 months is 0.543 kilograms.

Step-by-step explanation:

To find the amount of fluorine released into the air over 6 months, we first need to calculate the amount released in one week. The amount released per week is given as 47.97 g of C2H2F3Cl. To find the amount of fluorine released, we need to use the molar mass of C2H2F3Cl and the ratio of fluorine atoms to C2H2F3Cl molecules. The molar mass of C2H2F3Cl is 130.460 g/mol, and it contains 3 fluorine atoms per molecule. Therefore, the amount of fluorine released per week is (47.97 g / 130.460 g/mol) * 3 = 1.097 moles of fluorine.

To find the amount of fluorine released in 6 months, we multiply the amount released per week by the number of weeks in 6 months. There are approximately 26 weeks in 6 months. Therefore, the amount of fluorine released in 6 months is 1.097 moles/week * 26 weeks = 28.522 moles of fluorine.

To convert moles to kilograms, we need to multiply by the molar mass of fluorine (18.998 g/mol) and divide by 1000 to convert grams to kilograms. Therefore, the amount of fluorine released in 6 months is (28.522 moles * 18.998 g/mol) / 1000 = 0.543 kilograms of fluorine.

Answer 2

0.554 kg

Step-by-step 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