Answer: The mass of iron (III) oxide produced is 59.9 grams.
Explanation:
To calculate the number of moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) .....(1)
Given mass of iron = 112 g
Molar mass of iron = 55.84 g/mol
Putting values in equation 1, we get:
\(\text{Moles of iron}=\frac{112g}{55.84g/mol}=2mol\)
Given mass of oxygen gas = 24 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:
\(\text{Moles of oxygen gas}=\frac{24g}{32g/mol}=0.75mol\)
The given chemical equation follows:
\(4Fe+3O_2\rightarrow 2Fe_2O_3\)
By Stoichiometry of the reaction:
4 moles of iron reacts with 3 moles of oxygen gas
So, 0.75 moles of iron will react with = \(\frac{3}{4}\times 0.75=0.5625mol\) of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, iron is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
4 moles of iron produces 2 moles of iron (III) oxide
So, 0.75 moles of iron will produce = \(\frac{2}{4}\times 0.75=0.375mol\) of iron (III) oxide
Now, calculating the mass of iron (III) oxide by using equation 1:
Molar mass of iron (III) oxide = 159.7 g/mol
Moles of iron (III) oxide = 0.375 moles
Putting values in equation 1, we get:
\(0.375mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(0.375mol\times 159.7g/mol)=59.9g\)
Hence, the mass of iron (III) oxide produced is 59.9 grams.
