Respuesta :
Answer: 32.9 Liters
Explanation:
To calculate the moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]
1. moles of [tex]FeS_2=\frac{96.7g}{119.99g/mol}=0.806mol[/tex]
2. moles of [tex]O_2[/tex]
[tex]PV=nRT[/tex]
P = pressure of the gas = 1.20 atm
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]398K[/tex]
[tex]1.20\times 55.0=n\times 0.0821\times 398[/tex]
[tex]n=2.02[/tex]
[tex]4FeS_2(s)+11O_2(g)\rightarrow 2Fe_2O_3(s)+8SO_2(g)[/tex]
According to stoichiometry:
11 moles of oxygen reacts with 4 moles of [tex]FeS_2[/tex]
Thus 2.02 moles of oxygen reacts with =[tex]\frac{4}{11}\times 2.02=0.73[/tex] moles of [tex]FeS_2[/tex]
Thus oxygen acts as limiting reagent and [tex]FeS_2[/tex] is excess reagent.
As 11 moles of oxygen gives = 8 moles of [tex]SO_2[/tex]
2.02 moles of oxygen gives =[tex]\frac{8}{11}\times 2.02=1.47[/tex] moles of [tex]SO_2[/tex]
[tex]PV=nRT[/tex]
P = pressure of the gas = 1 atm (at STP)
R = Gas constant = [tex]0.0821\text{ L atm }mol^{-1}K^{-1}[/tex]
T = temperature of the gas = [tex]273K[/tex] (at STP)
[tex]1\times V=1.47\times 0.0821\times 273[/tex]
[tex]V=32.9L[/tex]
Thus volume of [tex]SO_2[/tex] (at STP) formed from the reaction is 32.9 L
The volume of SO₂ formed from the reaction at STP is 32.92 L
From the question,
We are to determine the volume of SO₂ formed from the reaction
The given balanced chemical equation for the reaction is
4FeS₂(s) + 11O₂(g) → 2Fe₂O₃(s) + 8SO₂(g)
This means,
4 moles of FeS₂ reacts with 11 moles of oxygen to produce 2 moles of Fe₂O₃ and 8 moles of SO₂
First, we will determine the number of moles of each reactant present
- For FeS₂
Mass = 96.7 g
From the formula
[tex]Number\ of\ moles =\frac{Mass}{Molar\ mass}[/tex]
Molar mass of FeS₂ = 119.99 g/mol
∴ Number of moles of FeS₂ present =[tex]\frac{96.7}{119.99}[/tex]
Number of moles of FeS₂ present = 0.8059 mole
- For O₂
Using the formula
PV = nRT
[tex]n =\frac{PV}{RT}[/tex]
Putting the given parameters into the formula, we get
[tex]n = \frac{1.2 \times 55.0}{0.08206 \times 398}[/tex]
[tex]n = \frac{66}{32.65988}[/tex]
n = 2.0208 moles
Since,
4 moles of FeS₂ reacts with 11 moles of oxygen to produce 8 moles of SO₂
Then,
[tex]\frac{2.0208 \times 4}{11}[/tex] moles of FeS₂ will react with 2.0208 moles of oxygen to produce [tex]\frac{2.0208 \times 4}{11} \times 2[/tex] moles of SO₂
That is,
0.7348 moles of FeS₂ will react with 2.0208 moles of oxygen to produce 1.4697 moles of SO₂
∴ Number of moles of SO₂ formed is 1.4697 moles
Now for the volume of SO₂ formed at STP
Since
1 mole of a gas has a volume of 22.4 L at STP
Then
1.4697 moles of the SO₂ will have a volume of 1.4697 × 22.4 L
1.4697 × 22.4 L = 32.92L
Hence, the volume of SO₂ formed from the reaction at STP is 32.92 L
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