A 185.00g sample of fluorine gas and 4 moles of xenon were placed in a flask at 0°C and 2.5 atm. Deduce volume of the gaseous mixture
Answers
Answer:
assume that 185.00 grams of fluorine gas and 4.0-moles of Xenon gas are contained in a flask at 0degree Celcius and 2.5-atm of pressure.
a) calculate the volume of the flask and the partial pressure of each gas
b) 23.00-gm of Li metal is introduced into the flask, and a violent reaction occurs in which one of the reactants is entirely consumed. what weight of LiF formed?
c) calculate the partial pressure of any gas(es) present after the reaction in part B is complete and the temperature has been brought back to 0 degree celcius. (the volume of solid reactants and products may be ignored)
The molecular mass of fluorine gas (F2) is 37.997 g/mol, so 185.00 g of F2 contains 185/37.997 = 4.8688 moles of F2.
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Answer:
- The molecular mass of fluorine gas (F2) is 37.997 g/mol, so 185.00 g of F2 contains 185/37.997 = 4.8688 moles of F2.
.Initially, then, the flask contains a total of 4 + 4.8688 = 8.8688 moles of gas.
If we assume both the F2 and Xe gases obey the ideal gas law, then we can use this law to calculate the volume of the flask.
We have that:
P*V = n*R*T
where P is the pressure
V is the volume
n is the number of moles
R is the unversal gas constant = 8.2057 * 10^-2 liter*atm/(mol*K)
T is the absolute (kelvin) temperature.
Plugging the values we have for this case, and solving for V:
V = (8.8688 mol * 8.2057 * 10^-2 liter*atm/(mol*K) * 273.15 K)/2.5 atm
V = 79.514 liters
The partial pressure of each gas is given by the mole fraction of that gas times the total pressure:
P_F2 = (4.8688 mol)/(8.8688 mol) * 2.5 atm = 1.372 atm
P_Xe = (4 mol)/(8.8688 mol) * 2.5 atm = 1.128 atm
(Note that the sum of the partial pressures must equal the total pressure, and as a check, 1.372 + 1.128 = 2.5).
If a piece of Li is introduced, it will react with the F2 gas (Xe is a "noble gas" or "intert gas", and only rarely forms stable compounds) to form LiF according to the reaction:
Li(s) + 0.5 F2(g) = LiF(s)
Each mole of Li that reacts consumes 1/2 mole of F2 gas.
We are told that 23 grams of Li are introduced. The atomic mass of Li is 6.9410 g/mol, so 23 grams contains 23/6.9410 = 3.3136 moles of Li. There are more moles of F present than there are Li, so all of the Li will react to form 3.3140 moles of LiF, leaving some excess F2 gas in the flask.
The molecular mass of LiF is 25.9394 gm/mol, so this amount of LiF will have a mass of 3.3140 mol * 25.9394 gm/mol = 85.9539 gm.
Each mole of LiF formed consumes half as many moles of F2 gas, so after the reaction, the flask will contain:
4.8688 moles - 3.3136/2 moles = 3.2120 moles of F2
We now have 3.2120 mol + 4 mol = 7.2120 moles of gas in the flask. We can now use this value, along with the volume of the flask that we calculated in the beginning in the ideal gas law to calculate the new total pressure in the flask, one it has cooled down to 273.15 K again (and neglecting the volume occupied by the solids):
P = 7.2120 mol * 8.2057 * 10^-2 liter*atm/(mol*K) * 273.15 K)/79.514 liters
P = 2.033 atm
The new partial pressures of each gas are given by:
P_F2 = (3.2120 mol / 7.2120 mol) * 2.033 atm = 0.905 atm
P_Xe = (4 mol / 7.2120 mol) * 2.033 atm = 1.128 atm