What mass of Aluminum hydroxide is produced when 50.0 ml of .200 M Al(NO3)3 reacts with 200ml of .100 M KOH?

Question

What mass of Aluminum hydroxide is produced when 50.0 ml of .200 M Al(NO3)3 reacts with 200ml of .100 M KOH?

Answer 1

balanced equation :

Al(NO3)3 + 3 KOH ------> Al(OH)3 + 3 KNO3

1mole of Al(NO3)3 reacts with 3 moles of KOH to produce 1 mole of Al(OH)3 & 3 moles of KNO3

so: moles of Al(NO3)3 = moles of Al(OH)3

moles of Al(NO3)3 = molarity X liters = 0.2 X 0.05 =0.01 mol

so:  moles of Al(OH)3 = 0.01 mol

mass = moles X molecular weight

molecular weight of Al(OH)3= 26.98 + [ (16+1) X 3 ]=77.98 g/mol

mass of Al(OH)3 = 0.01 X 77.98 = 0.7798 g