Question

A mixture of CS2(g) and excess O2(g) in a 10 L reaction vessel at 300K is under a pressure of 3 atm. When the mixture is ignited by a spark, it explodes. The vessel successfully contains the explosion, in which all of the CS2(g) reacts to give CO2 (g) and SO2(g). The vessel is cooled back to its original temperature of 300K, and the total pressure of the two product gases and the unreacted O2(g) is found to be 2.4 atm. Calculate the mass (in grams) of CS2(g) originally present.

Answer 1

Step-by-step explanation:

The reaction equation is as follows.

`CS_(2)(g) + 3O_(2)(g) \rightarrow CO_(2)(g) + 2SO_(2)(g)`

This means that 1 mole of
`CS_(2)` reacts with 3 moles of
`O_(2)` and gives 1 mole of
`CO_(2)` and 2 moles of
`SO_(2)`.

Now, let us assume that there are x moles of
`CS_(2)` and y moles of
`O_(2)`. According to the ideal gas equation, we have PV = nRT.

where, p = pressure, V = volume, n = no of moles, R = Universal gas constant, and T = temperature.

Expression for total pressure is as follows.

`p_(total) = p_{CS_(2)} + p_{O_(2)}`

=
`(xRT)/(V) + (yRT)/(V)`

=
`((x + y)RT)/(V)` ........ (1)

The given data is as follows.

`p_(total)` = 3 atm, R = 0.0821 L atm/mol K, T = 300 K, V = 10 L

Now, using equation (1)

x + y =
`(p_(total) * V)/(R * T)`

=
`(3 * 10)/(0.0821 * 300)` mol

= 1.218 mol ......... (2)

Again, when the reaction from x moles of
`CS_(2)`, x moles of
`CO_(2)` and 2x moles of
`SO_(2)` are produced. During this process 3x moles of
`O_(2)` are consumed as per the above reaction.

Therefore, after the reaction,

moles of
`CO_(2)` = x ,

moles of
`SO_(2)` = 2x ,

moles of
`O_(2)` = (y - 3x)

and,
`p_(total)` = 2.4 atm

Hence, again the value of
`p_(total)` is as follows.

`p_(total) = p_{CO_(2)} + p_{SO_(2)} + p_{O_(2)}`

=
`(xRT)/(V) + (2xRT)/(V) + ((y - 3x)RT)/(V)`

=
`((x + 2x + y -3x)RT)/(V)`

=
`(yRT)/(V)`

Hence, the value of y can be calculated as follows.

y =
`(p_(total) * V)/(RT)`

=
`(2.4 * 10)/(0.0821 * 300)` mol

= 0.974 mol .......... (3)

On solving both equations (2) and (3), we find out the value of x as 0.244 moles.

As the molecular weight of
`CS_(2)` is 76 g/mol. Now, calculate the mass of
`CS_(2)` originally present as follows.

`0.244 mol * 76 g/mol`

= 18.544 g

Thus, we can conclude that the mass (in grams) of
`CS_(2)(g)` originally present is 18.544 g.