<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:
- <u>For </u>
<u> :</u>
Given mass of = 352 g
Molar mass of = 32 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of and oxygen gas follows:
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:
Hence, the mass of iron (III) oxide produced is 782.5 grams