Lead(ll) nitrate and ammonium iodide react to form lead(lI) iodide and ammonium nitrate according to the reaction
Pb(NO3)2 (aq) + 2 NH4I (aq) →PbI2 (s) + 2 NH4NO3 (aq)
What volume (mL) of a 0.270 M NH4I solution is required to react with 173 mL of a 0.460 M Pb(NO3)2 solution?
How many moles of Pbl2 are formed from this reaction?
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