posted by Lisa on .
If 45.3 mL of liquid CS2 and 7.62 L of gaseous O2 measured at STP are reacted stoichiometrically according to the balanced equation, how many milliliters of liquid CS2 remain?
CS2(l) + 3O2(g) -> CO2(g) + 2SO2(g)
Molar Mass (g/mol)
The correct answer is 38.47437
I don't understand how to set up the problem. I believe I'm supposed to use the Ideal Gas Law at some point. But how? Can someone please give me a step by step explanation on how to solve a problem like this?
1.From the volume of CS2 calculate the mass of CS2 using the density.
2.Using the mass of CS2 and the molar mass of CS2 calculate the number of moles of CS2.
3. Assume that 1 mole of an ideal gas occupies 22.4 litres at STP, calculate the number of moles of O2. (this is by simple proportion).
4. From the question we are told that the CS2 is in excess, i.e. all the moles of O2 are used. So one third of the number of moles of O2 (from step 3) is the number of moles of CS2 used (from the proportions in the equation).
5. Subtract the number of moles of CS2 used (from step 4) from the total number of moles of CS2 (from step 1).
6. Now reverse step 2 and 1 using the number fo moles of CS2 remaining (from step 5) to calculate the volume of CS2 remaining.
[note that the molar mass of O2 is not needed in this calculation]