Chemistry, 21.06.2021 16:00, benjamenburton1

At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular chlorine with a total pressure of 4.00 atm. Calculate the following: a. How many moles of Ar are in the tank?
b. How many moles of Cl, are in the tank?
c. Total moles of gas in the tank.
d. The mole fraction of Ar.
e. The mole fraction of Cl2.
f. The Partial Pressure of Ar.
g. The Partial Pressure of Cl2.

Answers: 3

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Answers

answered: Loliii

For (a): The moles of Ar is 16.94 moles

For (b): The moles of Cl_2 is 16.94 moles

For (c): The total number of moles in a tank is 23.47 moles

For (d): The mole fraction of Ar is 0.722

For (e): The mole fraction of Cl_2 is 0.278

For (f): The partial pressure of Ar is 2.888 atm

For (g): The partial pressure of Cl_2 is 1.112 atm

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

For (a):

Given mass of Ar = 675.5 g

Molar mass of Ar = 39.95 g/mol

Plugging values in equation 1:

$\text{Moles of Ar}=\frac{675.5g}{39.95g/mol}=16.91 mol$

For (b):

Given mass of Cl_2 = 465.0 g

Molar mass of Cl_2 = 70.9 g/mol

Plugging values in equation 1:

$\text{Moles of }Cl_2=\frac{465.0g}{70.9g/mol}=6.56 mol$

For (c):

Total moles of gas in the tank = [16.91 + 6.56] mol = 23.47 mol

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .....(2)

where n is the number of moles

For (d):

Moles of Ar = 16.94 moles

Total moles of gas in the tank = 23.47 mol

Putting values in equation 2, we get:

$\chi_{Ar}=\frac{16.94}{23.47}\\\\\chi_{Ar}=0.722$

For (e):

Total mole fraction of the system is always 1

Mole fraction of Cl_2 = [1 - 0.722] = 0.278

Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture.

The equation for Raoult's law follows:

$p_A=\chi_A\times p_T$ .....(3)

where p_A is the partial pressure of component A in the mixture and p_T is the total partial pressure of the mixture

For (f):

We are given:

$\chi_{Ar}=0.722\\p_T=4.00atm$

Putting values in equation 3, we get:

$p_{Ar}=0.722\times 4.00atm\\\\p_{Ar}=2.888atm$

For (g):

We are given:

$\chi_{Cl_2}=0.278\\p_T=4.00atm$

Putting values in equation 3, we get:

$p_{Cl_2}=0.278\times 4.00atm\\\\p_{Cl_2}=1.112atm$

answered: Guest

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1

0.44 / 0.09 = 4.8

mgco3-5h2o"

answered: Guest

Where is the graph ?

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At 50.0 oC, a reinforced tank contains 675.5 grams of gaseous argon and 465.0 g of gaseous molecular...