The balanced equation for the reaction is as follows;
2H₂S + SO₂ —> 2H₂O + 3S
Stoichiometry of H₂S to SO₂ is 2:1
Limiting reactant is fully used up in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of H₂S moles - 8.0 g / 34 g/mol = 0.24 mol of H₂S
Number of SO₂ moles = 12.0 g / 64 g/mol = 0.188 mol of SO₂
According to molar ratio of 2:1
If we assume H₂S to be the limiting reactant
2 mol of H₂S reacts with 1 mol of SO₂
Therefore 0.24 mol of H₂S requires - 1/2 x 0.24 = 0.12 mol of SO₂
But 0.188 mol of SO₂ is present therefore SO₂ is in excess and H₂S is the limiting reactant.
H₂S is the limiting reactant
Amount of S produced depends on amount of H₂S present
Stoichiometry of H₂S to S is 2:3
2 mol of H₂S forms 3 mol of S
Therefore 0.24 mol of H₂S forms - 3/2 x 0.24 mol = 0.36 mol of S
Mass of S produced = 0.36 mol x 32 g/mol = 11.5 g of S is produced
My answer will be C. Law of Original lateral continuity. :)
Answer:
The answer to your question is False.
Explanation:
Data
1 mole of NH₄NO₃
0.75 moles of N₂O
Percent yield = 25%
Chemical reaction
NH₄NO₃ ⇒ N₂O + 2H₂O
Process
1.- Determine the theoretical yield
1 mol NH₄NO₃ ------------- 1 mol of N₂O
2.- Calculate the percent yield
Percent yield = Actual yield / Theoretical yield x 100
-Substitution
Percent yield = 0.75 / 1 x 100
-Simplification
Percent yield = 0.75 x 100
-Result
Percent yield = 75%
Conclusion
False, the actual percent yield is 75%
Answer:
Explanation:
¡Hola!
En este caso, al considerar la unidad de concentración de porcentaje masa/masa, podemos escribir su fórmula como:
Podemos identificar la masa de soluto (azúcar) como la incógnita, y resolverla como se muestra a continuación:
¡Saludos!
