Solution Stoichiometry
6. Chemical Quantities & Aqueous Reactions / Solution Stoichiometry / Problem 6
Lead (ll) nitrate and ammonium iodide react to form lead(lI) iodide and ammonium nitrate according to the following reaction:
Pb(NO3)2 (aq) + 2 NH4I (aq) → PbI2 (s) + 2 NH4NO3 (aq)
How many mL of a 0.178 M ammonium iodide solution is required to react with 232 mL of a 0.355 M Lead (ll) nitrate solution? And how many moles of lead(lI) iodide are formed from this reaction?
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