Question

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.

Answer 1

Answer: The correct answer is Option d.

Step-by-step explanation:

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For magnesium:

Given mass of magnesium = 41.0 g

Molar mass of magnesium = 24.3 g/mol

Putting values in equation 1, we get:

`\text{Moles of magnesium}=(41.0g)/(24.3g/mol)=1.69mol`

• For iron(III) chloride:

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:

`\text{Moles of iron(III) chloride}=(175g)/(162.2g/mol)=1.08mol`

The given chemical equation follows:

`3Mg(s)+2FeCl_3(s)\rightarrow 3MgCl_2(s)+2Fe(s)`

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 =
`(3)/(2)* 1.08=1.62mol` 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:

`0.07mol=\frac{\text{Mass of magnesium}}{24.3g/mol}\\\\\text{Mass of magnesium}=(0.07mol* 24.3g/mol)=1.7g`

Mass of excess reagent left = 1.7 grams

Hence, the correct answer is Option d.