A mixture of CS_2(g) and excess O_2(g) in a 10.0-L reaction vessel at 100.0 degr

Question

A mixture of CS_2(g) and excess O_2(g) in a 10.0-L reaction vessel at 100.0 degree C is under a pressure of 3.00 atm. When the mixture is ignited by a spark, it explodes. The vessel successfully contains the explosion, in which all of the CS_2(g) reacts to give CO_2(g) and SO_2(g). The vessel is cooled back to its original temperature of 100.0 degree C, and the pressure of the mixture of the two product gases and the unreacted O_2(g) is found to be 2.40 atm. Calculate the mass (in grams) of CS_2(g) originally present.

Explanation / Answer

In the following equations, the partial pressures of each gas will be represented by the compound formula in brackets.

4/3 * ([CO2] + [SO2]) + [O2] = 3.15 atm
This is the equation describing the pressure before the reaction.

[CO2] + [SO2] + [O2] = 2.50 atm
This is the equation describing the pressure after the reaction.

[SO2] = 2 [CO2]
This relates the pressure of CO2 to SO2.

Substituting equation 3 into equation 2, we get

3 [CO2] + [O2] = 2.50 atm

Solving for [O2] and plugging into equation 1, we get
4/3*3[CO2] + 2.50 - 3[CO2] = 3.15 atm

Solve for [CO2].
[CO2] = 0.65 atm

By the relations shown above, the remaining partial pressures are found to be.
[SO2] = 1.30 atm
[O2] = 0.55 atm

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