When the iron(III) oxide reacts with hydrochloric acid, iron (III) chloride and water are formed. How many gras of iron (III) chloride are formed from 10.0g of iron (III) oxide

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

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

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