# chemistry

What volume of a 0.610 M NH4I solution is required to react with 525 mL of a 0.300 M Pb(NO3)2 solution?

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1. Pb(NO3)2 + 2NH4I ==> PbI2 + 2NH4NO3

mols Pb(NO3)2 = M x L = estimated 0.16
mols NH4I needed = estd 2*0.16 = about 0.32
M NH4I = mols NH4I/L NH4I. You know mols and M, solve for L.

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