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how many ml. of .112 M Pb(NO3)2 are needed to completly react w/ 25.0ml of .105 M Kl? GIVEN: Pb(NO3)2(aq)+?

+2Kl(aq)>>>>PbI2(s)+2KNO3(aq)

2007-05-14 17:31:24 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry

1 answers

0.025*0.105 = 0.0002625 M KI
(0.0002625 M KI)/2 = 0.00013125 M Pb(NO3)2
(0.00013125 M Pb(NO3)2)/(0.112) = 1.171875 ml

2007-05-14 21:12:22 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋

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