Question

In an experiment, 57.4 g of iron (II) chloride solution reacts with 45.3 g of sodium hydroxide solution.

FeCla(an) + 3NaOH(an) -› Fe(OH) scs) + 3NaC|(ag)

Which reactant is the limiting reagent?

How much (in grams) of each product forms?

How much of the excess reagent (in grams) will be leftover after the reaction is complete?

Answer 1

FeCl₂ + 2NaOH = Fe(OH)₂ + 2NaCl

n(FeCl₂) = m(FeCl₂) / Mr(FeCl₂) = 57.4 g / 126.75 g/mol = 0.45 moles

n(NaOH) = m(NaOH) / 2Mr(NaOH) = 45.3 g / 2*40 g/mol = 0.57 moles

NaOH - excess reagent

FeCl₂ - limiting reagent

m(Fe(OH)₂) = (m(FeCl₂) * Mr(Fe(OH)₂) / Mr(FeCl₂) = (57.4 g * 89.86 g/mol) / 126.75 g/mol = 40.69 g

m(NaCl) = (m(FeCl₂) * 2Mr(NaCl) / Mr(FeCl₂) = ( 57.4 g * (2*58.44 g/mol)) / 126.75 g/mol = 52.93 g

m(NaOH) = (m(FeCl₂) * 2Mr(NaOH) / Mr(FeCl₂) = ( 57.4 g * (2*40 g/mol)) / 126.75 g/mol = 36.23 g

m(NaOH) _{after the reaction is complete} = 45.3 g - 36.23 g = 9.07 g