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According to the following reaction, what volume of 0.25 M CaCl2 solution is required to react exactly with 50.0 mL of 0.20 M Pb(NO3)2 solution? CaCl2(aq) + Pb(NO3) (aq)- PbCl2(s) + 2 KNO3(aq).
12 years ago

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Steve
50.0 mL of 0.20 M Pb(NO3)2 contains .05*.02 = .01 moles

according to the equation, each mole of PbNO3 requires one mole of CaCl2.

so, since 1L of .25M CaCl2 contains .25 moles, you want .01/.25 = .04L = 40mL
12 years ago

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