<u>Answer:</u> The correct answer is Option d.
<u>Explanation:</u>
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:
![\text{Moles of magnesium}=\frac{41.0g}{24.3g/mol}=1.69mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20magnesium%7D%3D%5Cfrac%7B41.0g%7D%7B24.3g%2Fmol%7D%3D1.69mol)
- <u>For iron(III) chloride:</u>
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}=\frac{175g}{162.2g/mol}=1.08mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20iron%28III%29%20chloride%7D%3D%5Cfrac%7B175g%7D%7B162.2g%2Fmol%7D%3D1.08mol)
The given chemical equation follows:
![3Mg(s)+2FeCl_3(s)\rightarrow 3MgCl_2(s)+2Fe(s)](https://tex.z-dn.net/?f=3Mg%28s%29%2B2FeCl_3%28s%29%5Crightarrow%203MgCl_2%28s%29%2B2Fe%28s%29)
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:
![0.07mol=\frac{\text{Mass of magnesium}}{24.3g/mol}\\\\\text{Mass of magnesium}=(0.07mol\times 24.3g/mol)=1.7g](https://tex.z-dn.net/?f=0.07mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20magnesium%7D%7D%7B24.3g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20magnesium%7D%3D%280.07mol%5Ctimes%2024.3g%2Fmol%29%3D1.7g)
Mass of excess reagent left = 1.7 grams
Hence, the correct answer is Option d.