Respuesta :
Answer: The mass of molten iron formed will be 1.92 kg
Explanation:
We are given:
Moles of iron (III) oxide = 17.2 moles
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of aluminium = 2.28 kg = 2280 g (Conversion factor: 1 kg = 1000 g)
Molar mass of aluminium = 27 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of aluminium}=\frac{2280g}{27g/mol}=84.44mol[/tex]
The chemical equation for the reaction of aluminium and iron (III) oxide follows:
[tex]2Al+Fe_2O_3\rightarrow Al_2O_3+2Fe[/tex]
By Stoichiometry of the reaction:
1 mole of iron (III) oxide reacts with 2 moles of aluminium
So, 17.2 moles of iron (III) oxide will react with = [tex]\frac{2}{1}\times 17.2=34.4mol[/tex] of aluminium
As, given amount of aluminium is more than the required amount. So, it is considered as an excess reagent.
Thus, iron (III) oxide is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of iron (III) oxide produces 2 moles of iron
So, 17.2 moles of iron (III) oxide will produce = [tex]\frac{2}{1}\times 17.2=34.4mol[/tex] of iron
Now, calculating the mass of iron by using equation 1, we get:
Molar mass of iron = 55.8 g/mol
Moles of iron = 34.4 moles
Putting values in equation 1, we get:
[tex]34.4mol=\frac{\text{Mass of iron}}{55.8g/mol}\\\\\text{Mass of iron}=(34.4mol\times 55.8g/mol)=1919.5g=1.92kg[/tex]
Hence, the mass of molten iron formed will be 1.92 kg
