Question

Magnesium reacts with iron(III) chloride to form magnesium chloride (which can be used in fireproofing wood and in disinfectants) and iron. A mixture of 43.5 g Mg and 247 g iron(III) chloride is allowed to react. What is the limiting reactant and what is the mass, in grams, of the excess reactant remains

Answer 1

Mg is the limiting reactant.

52.9 g of FeCl₃, remains after the reaction is complete.

Step-by-step explanation:

First of all, we state the reaction's equation:

3Mg (s) + 2FeCl₃ (aq) → 2Fe (s) + 3MgCl₂ (s)

We determine the mol of each reactant:

43.5 g . 1mol / 24.3g = 1.79 mol of Mg

247 g . 1mol/ 162.2g = 1.52 mol of Iron (III) chloride.

Ratio is 2:3. 2 mol of chloride need 3 moles of Mg to react,

Then 1.52 moles will react with (1.52 . 3) /2 = 2.29 moles

We have 1.79 moles of Mg and we need 2.29, so the limiting reactant is the Mg.

We confirm the chlorine as the excess reactant:

3 moles of Mg need 2 moles of FeCl₃ to react:

1.79 moles of Mg will react to (1.79 . 2) /3 = 1.19 moles of FeCl₃

We have 1.52 moles of FeCl₃ and we only need 1.19, so it is ok, we said that the FeCl₃ is the excess.

After the reaction goes complete, (1.52 - 1.19) moles of FeCl₃ remains.

1.52 - 1.19 = 0.33 moles. We convert them to mass:

0.33 mol . 162.2g /1mol = 52.9 g