Answer:
10.1 g of FeCl₃ are formed by the reaction
Explanation:
First step is to determine the reaction where the reactants are Fe₂O₃ and HCl in order to produce FeCl₃ and H₂O.
Equation is: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O
We assume the acid is in excess, so the limiting reagent will be the oxide.
Let's work with mass:
1 mol of Fe₂O₃ is 159.7 g
2 mol of FeCl₃ is 162.2 g
So now we propose a rule of three:
159.7 g of oxide can produce 162.2 grams of chloride
Then, 10 g of oxide will produce (10 . 162.2) / 159.7 = 10.1 g of FeCl₃
Answer:
20.3 grams of FeCl3 will be formed
Explanation:
Step 1: Data given
iron(III) oxide = Fe2O3
hydrochloric acid = HCl
iron (III) chloride = FeCl3
water = H2O
Mass of Fe2O3 = 10.0 grams
Molar mass Fe2O3 = 159.69 g/mol
Step 2: The balanced equation
Fe2O3 +6HCl → 2FeCl3 + 3H2O
Step 3: Calculate moles Fe2O3
Moles Fe2O3 = mass Fe2O3 / molar mass Fe2O3
Moles Fe2O3 = 10.0 grams/ 159.69 g/mol
Moles Fe2O3 = 0.0626 moles
Step 4: Calculate moles of FeCl3
For 1 mol Fe2O3 we need 6 moles HCl to produce 2 moles FeCl3 and 3 moles H2O
For 0.0626 moles Fe2O3 we'll have 2*0.0626 = 0.1252 moles FeCl3
Step 5: Calculate mass FeCl3
Mass FeCl3 = moles FeCl3 * molar mass FeCl3
Mass FeCl3 = 0.1252 moles * 162.2 g/mol
Mass FeCl3 = 20.3 grams
20.3 grams of FeCl3 will be formed
