What is the energy given off by the
1 answer:
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Answer:
Limiting reactant= SO₂, Theoretical yield = 9.6g SO₃
Explanation:
The first step is is to find the limiting reactant; you do this using stoichiometry.
The reactant that produces the smallest amount of sulfur trioxide SO₃, is the limiting reactant.
In this problem, SO₂ is the limiting reactant.
Now to find the theoretical yield we will use the smallest value of SO₃.
I have uploaded a picture to better illustrate the problem and hope it helps.
Q1)
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm
we can use the ideal gas law equation to find the total pressure of the system ;
PV = nRT
where P - pressure
V - volume - 7 x 10⁻³ m³
n - number of moles
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in K - 273 + 25 °C = 298 K
substituting the values in the equation
P x 7 x 10⁻³ m³ = 0.857 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 303.33 kPa
1 atm = 101.325 kPa
Therefore total pressure - 303.33 kPa / 101.325 kPa/atm = 2.99 atm
Q2)
partial pressure is the pressure exerted by the individual gases in the mixture.
partial pressure for each gas can be calculated by multiplying the total pressure by mole fraction of the individual gas.
total number of moles - 0.477 + 0.265 + 0.115 = 0.857 mol
mole fraction of He -
mole fraction of Ne -
mole fraction of Ar -
partial pressure - total pressure x mole fraction
partial pressure of He - 2.99 atm x 0.557 = 1.67 atm
partial pressure of Ne - 2.99 atm x 0.309 = 0.924 atm
partial pressure of Ar - 2.99 atm x 0.134 = 0.401 atm