Regarding the question I posted earlier (How many mL of an impure liquid CS2 mixture (density 1.26 gmL-1) which contains 93.0% of CS2 by mass, must be burned to obtain 12.8g SO2?)>>volume is 6.49 mL

what volume of pure oxygen at 25.0 degrees C and 100 kPa would be consumed (assume the carbon content burns to CO2).

Thanks!

Use PV = nRT

Remember T is Kelvin.
Use the mass CS2 and convert to moles for n. There are 101.325 kPa in 1 atm.

I can't seem to get the answer (7.43)

PV=nRT
100V= 0.107334995 (8.314) (298.15)

V= 2.66

Any ideas?

Rule #1 before we talk about where you went wrong. Don't copy all those digits on the calculator. Use 0.1073 if you wish but no more.

Now for the problem.
How did you get the 0.1073 moles.
I would have started with the 12.8 g SO2 which is 12.8/64 = 0.200 moles SO2. Now convert that to moles O2 (which you didn't do) using the coefficients in the balanced equation.
0.200 moles SO2 x (3 moles O2/2 moles SO2) = 0.300. Now plug that in. I get 7.436 liters which rounds to 7.44 L. I get 7.43 if I use 298 instead of 298.15 for T and 298.15 isn't justified.

To find the volume of pure oxygen that would be consumed when burning the impure liquid CS2 mixture, we need to first calculate the moles of CS2 and then use the balanced chemical equation to determine the stoichiometry of CS2 to oxygen.

1. Calculate the moles of CS2:
First, convert the mass of CS2 to moles using its molar mass.
Molar mass of CS2 = 12.01 g/mol (C) + 32.07 g/mol (S) x 2 = 76.14 g/mol

moles of CS2 = mass of CS2 / molar mass of CS2
moles of CS2 = 12.8 g / 76.14 g/mol = 0.1683 mol

2. Use the balanced chemical equation to determine the stoichiometry:
The balanced equation for the combustion of CS2 is:
CS2 + 3O2 → CO2 + 2SO2

From the equation, we can see that 1 mole of CS2 reacts with 3 moles of oxygen. Therefore, we need to calculate the moles of oxygen required.

moles of O2 = moles of CS2 x (3 moles of O2 / 1 mole of CS2)
moles of O2 = 0.1683 mol x (3/1) = 0.5049 mol

3. Convert moles of oxygen to volume using the ideal gas law:
Now, we need to convert moles of oxygen to volume using the ideal gas law:

PV = nRT

Where:
P = Pressure (in atm) = 100 kPa = 0.9869 atm (using the conversion 1 kPa = 0.9869 atm)
V = Volume (in liters)
n = moles of oxygen
R = Ideal gas constant = 0.0821 L.atm/(mol.K)
T = Temperature (in Kelvin) = 25.0°C + 273.15 = 298.15 K

V = nRT / P
V = 0.5049 mol x 0.0821 L.atm/(mol.K) x 298.15 K / 0.9869 atm
V ≈ 12.78 L

Therefore, the volume of pure oxygen that would be consumed is approximately 12.78 liters.