Respuesta :
Answer: The limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For magnesium:
Given mass of magnesium = 41.0 g
Molar mass of magnesium = 24 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of magnesium}=\frac{41.0g}{24g/mol}=1.708mol[/tex]
- For iron(III) chloride:
Given mass of iron(III) chloride = 175.0 g
Molar mass of iron(III) chloride = 162.2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of iron(III) chloride}=\frac{175g}{162.2g/mol}=1.708mol[/tex]
The chemical equation for the reaction of magnesium and iron(III) chloride follows:
[tex]3Mg+2FeCl_3\rightarrow 3MgCl_2+2Fe[/tex]
By Stoichiometry of the reaction:
3 moles of magnesium reacts with 2 moles of iron(III) chloride
So, 1.708 moles of magnesium will react with = [tex]\frac{2}{3}\times 1.708=1.114mol[/tex] of iron(III) chloride
As, given amount of iron(III) chloride is more than the required amount. So, it is considered as an excess reagent.
Thus, magnesium is considered as a limiting reagent because it limits the formation of product.
Moles of excess reactant left (iron(III) chloride) = [1.708 - 1.114] = 0.594 moles
Now, calculating the mass of iron(III) chloride from equation 1, we get:
Molar mass of iron(III) chloride = 162.2 g/mol
Moles of iron(III) chloride = 0.594 moles
Putting values in equation 1, we get:
[tex]0.594mol=\frac{\text{Mass of iron(III) chloride}}{162.2g/mol}\\\\\text{Mass of iron(III) chloride}=(0.594mol\times 162.2g/mol)=96.35g[/tex]
Hence, the limiting reactant is magnesium and mass of excess reactant present in the vessel is 96.35 grams.
