VERY URGENT! IB CHEMISTRY 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.