0.0810 g of a group 2 metal iodide, MI₂, was dissolved in water and made up to a total volume of 25.00 cm³.
Excess lead(II) nitrate solution (Pb(NO₃)₂(aq)) was added to the MI₂ solution to form a precipitate of lead(II) iodide (PbI₂). The precipitate was dried and weighed and it was found that 0.1270 g of precipitate was obtained.
a. Determine the number of moles of lead iodide formed.
b. Write an equation for the reaction that occurs.
c. Determine the number of moles of MI2 that reacted.
d. Determine the identity of the metal, M.