a)I obtained 1166.5 which would round to 1166 g (but the 485 limits us to 3 sig figures).
b)right on both.
c)I obtained 740.77 g chlorobenzene used which gives 401.22 remaining. The difference may be in rounding. I left all the numbers in my calculator as I went from one step to the other. I think your number is close enough that you probably worked the problem correctly. Of course we would get slightly different numbers for percent yield; I used your numbers and 50.306%
which will round to 50.31 but we aren't allowed that many significant figures anyway. Good work.
2C6H5Cl(chlorobenzene)+C2HOCl3(chloral)=C14H9Cl5(DDT)+H2O
in a gov. lab 1142 g of chlorobenzene is reacted with 485 g of chloral.
a). what mass of DDT is formed? 1166.179 g of C14H9Cl5
b). which reactant is limiting?C2HOCl3 which reactant is in excess?C6H5Cl
c). what mass of the excess reactant is left over? 397.56g C6H5Cl
d). if the actual yield is 200.0 g, what is the percent yield? 50.30%
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