Aqueous lithium hydroxide reacts with aqueous aluminum nitrate. if the reacting solutions contain 120.0 grams of reactant, how much precipitate is made, theoretically, and what mass of excess reactant remains?

The moles of each reagent present will be calcualted as

Moles = mass / molar mass

a) Mass of LiOH given = 120 g

Molas mass of LiOH = 24 g / mol

Moles of LiOH = 120 / 24 = 5 moles

b) mass of  Al(NO3)3 = 120 g

Molar mass = 213 g/mol

moles of Al(NO3)3 = 120 g / 213 = 0.56

The reaction between lithium hydroxide and aluminium nitrate will be

3LiOH (aq)  + Al(NO3)3(aq)   > 3LiNO3(aq)  + Al(OH)3 (s)

So here for three moles of LiOH we need one mole of Al(NO3)3 to form one mole of Al(OH)3

So with 5 moles of LiOH we need = 5/3 moles of Al(NO3)3 = 1.67 moles o Al(NO3)3

However we have only 0.56 moles of Al(NO3)3

Thus here Al(NO3)3 is the limiting reagent

1 mole of Al(NO3)3 will react with three moles of LiOH to give one mole of  Al(OH)3

0.56 moles of Al(NO3)3 will react with 0.56 X 3 moles of LiOH to give 0.56 moles of Al(OH)3

Mass of Al(OH)3  formed = moles X molar mass = 0.56 X 78 = 43.68 g

Moles of LiOH left = 5 - (3*0.56) = 3.32 moles

Mass of LiOH left = moles X molar mass = 3.32 X 24 = 79.68 g

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