9. Solutions
Solution Stoichiometry
9. Solutions Solution Stoichiometry
2PRACTICE PROBLEM
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?
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?