Question

Chlorine gas is held in a container with volume of 1.15L. The container initially held 3.53 x 10^−2 mol of chlorine gas, but after some of the gas was removed to carry out a reaction, the volume decreased to 0.790L while the temperature and pressure remained constant. How many grams of chlorine gas were removed from the container? The molar mass of chlorine gas (Cl2) is 70.9 g/mol. Give your answer in three significant figures.

Answer 1

Answer: 0.783 g

Step-by-step explanation:

N2= V2 X N1 / V2

N2 = 0.790 L X 3.53 X 10 ^-2 mol / 1.15 L

= 2.45 X 10^-2

N1-N2= 3.53 X 10^-2 - 2.425 X 10 ^-2

= 1.105 X 10^-2

Mchlorine = 1.105 X 10^-2 mol X 70.9 g/mol

= 0.783 g

Answer 2

The grams that were removed are 0.787

Step-by-step explanation:

We can raise two situations, with Ideal Gases Law.

First (S1): P. 1.15L =0.0353 . R .T

Second (S2): P. 0.790L = n . R .T

If we compare them, we must make a division, so we can know the moles of gas, that are still remained in the container.

S1/S2

P . 1.15L / P. 0.790L = 0.0353 . R.T / n . R. T

Pressure, temperature and R, are constant, so they can be cancelled.

So finally we have.

1.15L / 0.790L = 0.0353 moles / n

n = 0.0353 moles (0.790L /1.15L)

n = 0.0242 moles

Initially we have 0.0353 moles; afterwards we have 0.0242 moles. In the reaction we lost (0.0353 - 0.0242) = 0.0111 moles

The grams that were removed are:

Moles . molar mass = mass

0.0111 m . 70.9 g/m = 0.787 g