Answer :
Answer: The mass of iron (III) oxide produced is 782.5 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 [tex]FeS_2[/tex] :
Given mass of [tex]FeS_2[/tex] = 588 g
Molar mass of [tex]FeS_2[/tex] = 120 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }FeS_2=\frac{588g}{120g/mol}=4.9mol[/tex]
- For [tex]O_2[/tex] :
Given mass of [tex]O_2[/tex] = 352 g
Molar mass of [tex]O_2[/tex] = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of }O_2=\frac{352g}{32g/mol}=11mol[/tex]
The chemical equation for the reaction of [tex]FeS_2[/tex] and oxygen gas follows:
[tex]FeS_2+O_2\rightarrow Fe_2O_3+SO_2[/tex]
By Stoichiometry of the reaction:
1 mole of [tex]FeS_2[/tex] reacts with 1 mole of oxygen gas
So, 4.9 moles of [tex]FeS_2[/tex] will react with = [tex]\frac{1}{1}\times 4.9=4.9mol[/tex] of oxygen gas
As, given amount of oxygen gas is more than the required amount. So, it is considered as an excess reagent.
Thus, [tex]FeS_2[/tex] is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
1 mole of [tex]FeS_2[/tex] produces 1 mole of iron (III) oxide
So, 4.9 moles of [tex]FeS_2[/tex] will produce = [tex]\frac{1}{1}\times 4.9=4.9moles[/tex] 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:
[tex]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[/tex]
Hence, the mass of iron (III) oxide produced is 782.5 grams
The mass of Fe2O3 produced is 784 grams
Let's write the balance chemical reaction
Balanced Chemical reaction:
- FeS₂ + O₂ → Fe₂O₃ + SO₂
Therefore, the limiting reagent is FeS₂. The limiting reagent determines the amount of product formed. Therefore,
molar mass of FeS₂ = 56 + 64 = 120 g
molar mass of Fe₂O₃ = 112 + 48 = 160 g
Therefore,
120 g of FeS₂ produced 160 g of Fe₂O₃
588 g of FeS₂ will produce ?
cross multiply
mass of Fe₂O₃ produced = 160 × 588 / 120
mass of Fe₂O₃ produced = 94080 / 120
mass of Fe₂O₃ produced = 784 grams
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