1.
162g of solid aluminum reacts with 331g of aqueous lead (II) nitrate to produce Predict the
products. Then, write the balanced chemical equation: Identify the type of reaction: Determine
the limiting reactant : ??????????? ??? Calculate the amount (g) of aluminum nitrate formed:
Calculate the amount (g) of reactant in excess that remains:
Solution
2Al(s) + 3Pb(NO3)2(aq) ----------->2 Al(NO3)3(aq) +3Pb(s)
no of moles of Al   = W/G.A.Wt
= 162/27Â Â = 6 moles
no of moles of Pb(NO3)2Â Â = W/G.M.Wt
= 331/331Â Â = 1moles
2 moles of Al react with 3 moles of Pb(NO3)2
6 moles of Al react with = 3*6/2Â Â = 9 moles of Pb(NO3)2 is required.
Pb(NO3)2 is limiting reactant
3 moles Pb(NO3)2 react with Al to gives 2 moles of Al(NO3)3
1 moles of Pb(NO3)2 react with Al to gives = 2*1/3Â Â = 0.67 moles of Al(NO3)3
mass of Al(NO3)3 = no of moles * gram molar mass
= 0.67*213Â Â = 142.71g
3 moles of Pb(NO3)2 react with 2 moles of Al
1 moles of Pb(NO3)2 react with = 2*1/3Â Â Â = 0.67 moles of Al

2.
The no of moles of excess reactant remains after complete the reaction = 6-0.67Â Â = 5.33
moles of Al
The amount of excess reactant remains after complete the reaction = 5.33*27Â Â = 143.91g of
Al