A 0.810 g sample of impure CrCl3 was dissolved in water and required 37.0 mL of 0.125 M AgNO3 solution to react with the CrCl3 in the sample. What was the mass percent of CrCl3 in the sample?
I think the balances version would be
CrCl3(s) + 3AgNO3 (aq)--> 3AgCl(s) + Cr(NO3)3(aq)
but that's as far as I can get.
1
Expert's answer
2012-03-01T09:44:41-0500
in 1000 ml ----------0.125 moles in 37.0----------------x
x = 0.004625 moles
x 0.004625 CrCl3(s) + 3AgNO3 (aq)--> 3AgCl(s) + Cr(NO3)3(aq)
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