Answer to Question #318790 in General Chemistry for Gigi

Question

According to the following reaction, how many grams of water will be formed upon the complete reaction of 23.3 grams of iron(III) oxide with excess hydrochloric acid?

6HCl (aq) + Fe₂O₃ (s) 3H₂O (l) + 2FeCl₃ (aq)

Answer 1

6HCl (aq) + Fe₂O₃ (s) --->3H₂O (l) + 2FeCl₃ (aq)

1 mol Fe₂O₃ = 3 mol H₂O

i.e. 3 mol of H₂O will be produced when 1 mol of Fe₂O₃ reacted with excess HCl

1 mol Fe₂O₃ = 3 mol H₂O

molecular weight of Fe₂O₃ = 159.69 gm

molecular weight of H₂O = 18 gm

Convert into gram by multiplying molecular weight

1 mol Fe₂O₃ = 3 mol H₂O

Convert into gram

159.69 gram = 3×18 gram

159.69 gram = 54 gram

So how many gram of H₂O wii be produced when 23.3 gram of iron oxide get reacted with excess HCl

Gram of H₂O produced

= [(54 gm H2O)/(159.69 gm Fe₂O₃)]× (23.3 gmFe₂O₃ reacted)

= 7.879 gram H₂O produced ans.