Question

a)If a solution containing 119.31 g of silver nitrate is allowed to react completely with a solution containing 10.37g of lithium iodide, how many grams of solid precipitate will be formed?

b) How many grams of reactant in excess will remain after the reaction?

c) Assuming complete precipitation, how many moles of each ion remian in solution? If an ion is no longer in solution, enter zero for the number of moles.

Answer 1

The reaction between AgNO3(aq) and LiI(aq) can be written as

AgNO3(aq)+LiI(aq)→LiNO3(aq)+AgI(s), so the reaction gives rise to a precipitate of AgI

the reaction suggests that 1 mole of AgNO3 reacts with 1 mole of LiI to give AgI

molar ratio of AgNO3 :LiI ( theoretical) = 1:!

moles= mass/molar mass

Molar mass data : AgNO3= 170 g/mole, LiI= 134 g/mole and AgI=235

moles given : AgNO3 = 119.31/170 =0.7018, LiI= 10.37/134=0.077

actual molar ratio of AgNO3: LiI= 0.7018:0.077 : 0.7018/0.077 :1 =9.1:1

so excess is AgNO3 and all the LiI reacts . 1 mole of LiI theoretically gives 1 mole of AgI

0.077 moles of LII gives 0.077 moles of AgI. mas of AgI formed =moles* molar mass =0.077* 235=18.17 gm

moles of LiNO3 fomed remain in solution =0.077, moles of AgNO3 remaining = 0.7018-0.077 =0.6244

moles of Li+ in solution =0.077 and moles of NO3- from LiNO3=0.077

moles of Ag+ in solution =0.6244 and moles of NO3= 0.6244, total of NO3- in solution =0.6244+0.077= 0.7018