suppose 10 g of sulfur is reacted with 48 g of fluorine gas to make SF6, and after the reaction is complete, the temperature in the flask is 125 C.
Find the partial pressure of SF6 in the flask and the mole fraction of SF6 in the 6L flack after the reaction is complete.
S + 3F2 ==> SF6
moles S = 10/32.066 = ??
moles F2 = 48/38 = ??
S is used completely, 0.312 mole SF6 is formed and moles F2-moles used = moles F2 remaining un-reacted.
Use PV = nRT to solve for partial pressure SF6 and again for partial pressure F2 un-reacted.
moles fraction SF6 = moles SF6/total moles.
To find the partial pressure of SF6 in the flask, we need to apply the ideal gas law, which states:
PV = nRT
where:
P = pressure of the gas
V = volume of the flask
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin
First, let's calculate the number of moles of SF6. We can find this using the molar mass of SF6:
Molar mass of SF6 = 32.06 g/mol + (6 × 19.00 g/mol) = 146.06 g/mol
moles of SF6 = mass of SF6 / molar mass of SF6
= 10 g / 146.06 g/mol
≈ 0.0685 mol
Next, let's convert the temperature from Celsius to Kelvin:
T(K) = T(C) + 273.15
= 125 + 273.15
= 398.15 K
Now we can plug these values into the ideal gas law equation:
PV = nRT
P * 6 L = 0.0685 mol * 0.0821 L·atm/(mol·K) * 398.15 K
Solving for P, we get:
P = (0.0685 mol * 0.0821 L·atm/(mol·K) * 398.15 K) / 6 L
P ≈ 2.85 atm
Therefore, the partial pressure of SF6 in the flask is approximately 2.85 atm.
To find the mole fraction of SF6 in the 6L flask, we need to calculate the total number of moles of gas in the flask. This includes the moles of SF6 and the moles of any other gases present.
moles of F2 = mass of F2 / molar mass of F2
= 48 g / 38.00 g/mol
= 1.26 mol
Total moles of gas = moles of SF6 + moles of F2
= 0.0685 mol + 1.26 mol
= 1.3285 mol
Now we can calculate the mole fraction of SF6:
Mole fraction of SF6 = moles of SF6 / total moles of gas
= 0.0685 mol / 1.3285 mol
≈ 0.0516
Therefore, the mole fraction of SF6 in the 6L flask after the reaction is complete is approximately 0.0516.