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
