posted by lynne on .
I have been staring at this one for awhile now... I would so appreciate any help on approaching this problem!
A certain sample of coal contains 1.6 percent sulfur by mass. When the coal is burned, the sulfur is converted to sulfur dioxide. To prevent air pollution, this sulfur dioxide is treated with calcium oxide (CaO) to form calcium sulfate (CaSO3.) Calculate the daily mass (in kg) of CaO needed by a power plant that uses 6.60 x 10^6 kg of coal per day.
Do you mean calcium sulfite (CaSO3) or calcium sulfate(CaSO4)?
Your question states calcium sulfate but you have written calcium sulfite for the formula. I suspect that is a typo and you intend for it to be CaSO4.
Assuming you made a typo and it is CaSO4, calcium sulfate.
Convert 6.60 x 10^6 kg to S.
6.60 x 10^6 kg coal x 0.016 = ?? kg S.
There is 1 mol S in 1 mol SO2 and it requires 1 mol CaO to react with it. Therefore, you need not go through all of the intermediate steps; i.e., you do not need to calculate g SO2 formed and g CaO needed.Simply convert kg S to g S, convert that to mols S, convert that to mols CaO, convert to grams CaO and that to kg CaO.
Post your work if you get stuck.
This person meant sulfite.
just do what dr guy said cuz all is one to one ratio. foo
3293 kg CaO