We now consider two moles of ideal gas at the same initial state (3.0 l at 273 k). this time, we will first perform an isobaric compression, then an isothermal expansion to bring the gas to the final state with the same volume of 3.0 l, and at temperature 90.0 k. hint for part (a), solve the ideal gas law for the volume after the compression, and use this value to find the work done. note that you are given the initial temperature and the number of moles. for part (b), use the volume you calculated for part (a) and the final temperature in the equation for work for an isothermal process. click the hint button again to remove this hint. (a) how much work (in j) is done on the gas during the isobaric compression? wcompression = j (b) how much work (in j) is done on the gas during the isothermal expansion?

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.

i don't know people like to fight everywhere rip.

the correct answer to this question is 2