Note that one mole of the sulfate ion reacts with one mole of the barium ion.
Number of moles of product (= number of moles of barium ion) =
5.0 g/ (RMM of barium sulfate)
5.0/233.40 = 0.0214 moles
[note that this is less than
25 x 1.0/1000, the number of moles of barium ions present]
number of moles of sodium sulfate (= number of moles of sulfate ion)is
M x V/1000
3.0 x V/1000 = 0.0214
so V= 7.1 ml to two sig figs.
What volume of 3.0M Na2SO4 must be added to 25ml of 1.0M BaCl2 to produce 5.0g of BaSO4?
The answer is 7.1ml but I don't get it very well.
I thought I could use mole ratio of the balanced equation, but it did not work.
Thanks a lot!
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