Muriatic acid, HCl, is often used to remove rust. A solution of muriatic acid, HCl, reacts with Fe2O3 deposits on industrial equipment.
How many liters of 5.50 M HCl would be needed to react completely with 439 g Fe2O3? (molar mass of HCl = 36.46 g/mol) (molar mass of Fe2O3 =159.70 g/mol).
Use the equation:
Fe2O3(s) + 6HCl(aq) → 2FeCl3(aq) + 3H2O,
to solve this problem.

3L

Explanation:

Step 1:

The balanced equation for the reaction.

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

Step 2 :

Determination of the masses of HCl and Fe2O3 that reacted from the balanced equation. This is illustrated below:

Molar mass of Fe2O3 = 159.70g/mol

Molar mass of HCl = 36.46 g/mol

Mass of HCl from the balanced equation = 6 x 36.46 = 218.76g

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl

Step 3:

Determination of the mass of HCl needed to react with 439g of Fe2O3. This is illustrated below:

From the balanced equation above,

159.70g of Fe2O3 reacted with 218.76g of HCl.

Therefore, 439g of Fe2O3 will react with = (439 x 218.76) /159.70 = 601.35g of HCl.

Step 4:

Conversion of 601.35g of HCl to mole. This is illustrated below:

Molar mass of HCl = 36.46 g/mol

Mass of HCl = 601.35g

Number of mole = Mass/Molar Mass

Number of mole of HCl = 601.35/36.46

Number of mole of HCl = 16.49 moles

Step 5:

Determination of the volume of the HCl that reacted.

This is illustrated below:

Mole of HCl = 16.49 moles

Molarity of HCl = 5.50 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 16.49/5.5

Volume of HCl = 3L

Therefore the volume of HCl needed for the reaction is 3L

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