<u>Answer:</u> The mass of iron (III) oxide produced is 782.5 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
- <u>For </u>
<u> :</u>
Given mass of
= 588 g
Molar mass of
= 120 g/mol
Putting values in equation 1, we get:
![\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DFeS_2%3D%5Cfrac%7B588g%7D%7B120g%2Fmol%7D%3D4.9mol)
- <u>For </u>
<u> :</u>
Given mass of
= 352 g
Molar mass of
= 32 g/mol
Putting values in equation 1, we get:
![\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20%7DO_2%3D%5Cfrac%7B352g%7D%7B32g%2Fmol%7D%3D11mol)
The chemical equation for the reaction of
and oxygen gas follows:
![FeS_2+O_2\rightarrow Fe_2O_3+SO_2](https://tex.z-dn.net/?f=FeS_2%2BO_2%5Crightarrow%20Fe_2O_3%2BSO_2)
By Stoichiometry of the reaction:
1 mole of
reacts with 1 mole of oxygen gas
So, 4.9 moles of
will react with =
of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus,
is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of
produces 1 mole of iron (III) oxide
So, 4.9 moles of
will produce =
of iron (III) oxide
Now, calculating the mass of iron (III) oxide from equation 1, we get:
Molar mass of iron (III) oxide = 159.7 g/mol
Moles of iron (III) oxide = 4.9 moles
Putting values in equation 1, we get:
![4.9mol=\frac{\text{Mass of iron (III) oxide}}{159.7g/mol}\\\\\text{Mass of iron (III) oxide}=(4.9mol\times 159.7g/mol)=782.5g](https://tex.z-dn.net/?f=4.9mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20iron%20%28III%29%20oxide%7D%7D%7B159.7g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20iron%20%28III%29%20oxide%7D%3D%284.9mol%5Ctimes%20159.7g%2Fmol%29%3D782.5g)
Hence, the mass of iron (III) oxide produced is 782.5 grams