Question

Which is the limiting reactant when 15.0 g of iron (III) sulfate reacts with an equal mass of sodium hydroxide

Answer 1

Moles of Iron iii sulphate present = 15g/399.88gmol-1

=0.0375mol

Moles of NaoH present = 15g/39.997gmol-1

=0.375mol

According to balanced equation, Mole ratios of Iron iii sulfate : NaOH = 1:6

Therefore Moles of NaOH that is required to react with Iron iii sulfate =6×0.0375 = 0.225mol

Thus NaOH is in excess

And it follows that Iron iii sulfate is therefore the limiting reagent