Magnesium (used in the manufacture of light alloys) reacts with iron(iii) chloride to form magnesium chloride and iron. 3mg(s) + 2fecl₃(s) → 3mgcl₂(s) + 2fe(s)a mixture of 41.0 g of magnesium ( = 24.31 g/mol) and 175 g of iron(iii) chloride ( = 162.2 g/mol) is allowed to react. identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete. a) limiting reactant is mg: 67g of fecl₃ remains. b) limiting reactant is mg: 134 g fecl₃ remains. c) limiting reactant is mg: 104 g fecl₃ remains. d) limiting reactant is fecl₃: 1.7 g of mg remans. e) limiting reactant is fecl₃: 87.2 g of mg remains.

The correct answer is Option d.

Explanation:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24.3 g/mol

Putting values in equation 1, we get:

Given mass of iron(III) chloride = 175 g

Molar mass of iron(III) chloride = 162.2 g/mol

Putting values in equation 1, we get:

The given chemical equation follows:

By Stoichiometry of the reaction:

2 moles of iron(III) chloride reacts with 3 moles of magnesium

So, 1.08 moles of iron(III) chloride will react with = of magnesium

As, given amount of magnesium metal is more than the required amount. So, it is considered as an excess reagent.

Thus, iron(III) chloride is considered as a limiting reagent because it limits the formation of product.

Moles of excess reagent (magnesium) left = 1.69 - 1.62 = 0.07 moles

Now, calculating the mass of excess reagent by using equation 1, we get:

Molar mass of magnesium = 24.3 g/mol

Moles of magnesium = 0.07 moles

Putting values in equation 1, we get:

Mass of excess reagent left = 1.7 grams

Hence, the correct answer is Option d.

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