The theoretical yield of H₂S is 13.5 g.
The percent yield is 75.5 %.
<h3>What is the theoretical yield of H₂S from the reaction?</h3>
The equation of the reaction is given below:
Moles of FeS reacting = mass/molar mass
Molar mass of FeS = 88 g/mol
Moles of FeS reacting = 35/88 = 0.398 moles
Moles of H₂S produced = 0.398 moles
Molar mass of H₂S = 34 g/mol
Mass of H₂S produced = 0.398 * 34 = 13.5 g
Theoretical yield of H₂S is 13.5 g.
- Percent yield = actual yield/theoretical yield * 100%
Actual yield of H₂S = 10.2 g
Percent yield = 10.2/13.5 * 100%
Percent yield = 75.5 %
In conclusion, the actual yield is less than the theoretical yield.
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Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
____________________________
2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.
<u>Explanation:</u>
For the given nuclear reaction:
We are given:
Mass of 
= 39.963998 u
Mass of 
= 39.962591 u
To calculate the mass defect, we use the equation:
Putting values in above equation, we get:
To calculate the energy released, we use the equation:
(Conversion factor: 
)
Hence, the energy released in the given nuclear reaction is 1.3106 MeV.
Number 2 lower entropy and higher entropy
