[sami7ronaldo]

Hi guys ... I really need your immediate help with this assignment for partial and vapor pressure ...
Can any one solve these two questions with providing the explanation . I really need the answers >>
The 2 questions :

1)   10 g H2O is left in a closed-container with 5 L volume at 25oC. (dwater = 1 g/cm3)

a)   The liquid water is collected after one week. How much water is obtained?
b)   The volume is increased to 20 L. How much liquid water is obtained after one week?
c)   What should be the volume of the container for all the liquid water to evaporate?






2)   116 g butane and 45 g O2 is left in a 5 L container. After the reaction, the temperature of the container is adjusted to 200oC.

a)   What is the partial pressure of CO2 at 200oC?
b)   What is the partial pressure of CO2 at 25oC?


I hope you can solve them as soon as possible ... thanks

[sami7ronaldo]

Hi here is my attempts :
1) 10 g H2O is left in a closed-container with 5 L volume at 25oC. (dwater = 1 g/cm3)

a) The liquid water is collected after one week. How much water is obtained?

Solution:

Vapour Pressure at 25C = 23.76 mmHg = 23.76/ 760 = 0.0313 atm

number of moles of Water Vapour = PV / TR = (0.0313 * 5) / (0.0812 * (25+273)) = 6.46 * 10^-3 moles

Grams of liquid water = 10 - 18(6.45 * 10^-1) = 9.88 grams




2) 116 g butane and 45 g O2 is left in a 5 L container. After the reaction, the temperature of the container is adjusted to 200C.

a) What is the partial pressure of CO2 at 200C?

Solution:
2C4H10 (l) + 13O2 (g) -> 10H2O (l) + 8CO2 (g)

Moles of 02 = 45 / (16*2) = 1.4 mole
Moles of C4H10 = 116 / (4*12 + 10) = 2
O2 is the limiting reagent

Moles of CO2 = 8/13 * 1.4 = 0.86 moles
Partial pressure of C02 = (0.86 * 0.0821 * (200+273)) / 5 = 6.7 atm



I would like confirmation or correction on my solutions.
And a little help with the rest >>>

[Borek]

So far you did everything correctly, and the rest is not much different. Just don't round down intermediate results - report them as rounded, but use them in calculations with full accuracy (or at least with 2 or 3 so called "guard digits").

[sami7ronaldo]

1-b- n=PV/RT
n=0.0313x20/0.082x(25+273.15) = 0.026 moles of H2O
10-(0.026x18) = 9.532 g H2O

1-c- moles of H2O in the container = 10/18 = 0.55 moles
For all water to evaporate , the pressure must be 760 mmHg = 1 atm
and the temperature must be 100 Celsius
V= nRT / P = 0.55x0.082x(100+273.15) / 1 = 16.8 L 
___________________________________________
2-b- Partial pressure of CO2 at 25oC = 0.86 x 0.82 x (25+273.15) / 5 = 4.2 atm
___________________________________________
Here is the rest of the solutions >>> please confirm me that the answers are correct . if not correct me please . THX

[sjb]

2) 116 g butane and 45 g O2 is left in a 5 L container. After the reaction, the temperature of the container is adjusted to 200C.

a) What is the partial pressure of CO2 at 200C?

Solution:
2C4H10 (l) + 13O2 (g) -> 10H2O (l) + 8CO2 (g)

Moles of 02 = 45 / (16*2) = 1.4 mole
Moles of C4H10 = 116 / (4*12 + 10) = 2
O2 is the limiting reagent

Moles of CO2 = 8/13 * 1.4 = 0.86 moles
Partial pressure of C02 = (0.86 * 0.0821 * (200+273)) / 5 = 6.7 atm



I would like confirmation or correction on my solutions.
And a little help with the rest >>>

So if there is 0.86 mol of CO2, is there any other gas left in the container?

[mugabo daniel]

since oxygen is the limiting factor, it will react with only 0.215 moles of butane leaving 1.785 moles.
   13 moles of oxygen react with 2 moles of butane
   
    moles of butane used = (2/13)*1.4=0.215 moles
    moles of unused butane = 2-0.215 =1.785moles.
At 200, water vapour exists.