Determine the percent yield for the reaction between 118 g of Sb2S3 and excess oxygen if 74 g of Sb4O6 is recovered along with an unknown amount of sulfur dioxide gas (include 2 decimal places in your answer).
___ Sb2S3 + ___ O2 ⟶ ___ Sb4O6 + ___ SO2
2Sb2S3 + 9O2 ⟶ Sb4O6 + 6SO2
mols Sb2S3 initially = grams/molar mass = ?
mols Sb4O6 formed = 1/2 mols Sb2S3 initially.
grams Sb4O6 = mols Sb4O6 x molar mass Sb4O6 = theoretical yield = TY
Actual yield from the problem = AY = 74 g.
% yield = (AY/TY)*100 ==> ?
Post your work if you get stuck.
To determine the percent yield, we need to compare the amount of the desired product that was actually obtained (in this case, Sb4O6) to the amount that would be obtained if the reaction had gone to completion. The balanced equation for the reaction is:
3 Sb2S3 + 9 O2 ⟶ 2 Sb4O6 + 6 SO2
From the given information, we know that 118 g of Sb2S3 was reacted and 74 g of Sb4O6 was recovered. We need to determine the theoretical yield of Sb4O6.
First, let's calculate the molar masses of Sb2S3 and Sb4O6:
Molar mass of Sb2S3:
2 × (atomic mass of Sb) + 3 × (atomic mass of S) =
2 × (121.75 g/mol) + 3 × (32.06 g/mol) = 339.82 g/mol
Molar mass of Sb4O6:
4 × (atomic mass of Sb) + 6 × (atomic mass of O) =
4 × (121.75 g/mol) + 6 × (16.00 g/mol) = 747.06 g/mol
Next, we can use the molar masses to convert the mass of Sb2S3 to moles of Sb2S3:
Moles of Sb2S3 = Mass of Sb2S3 / Molar mass of Sb2S3
Moles of Sb2S3 = 118 g / 339.82 g/mol ≈ 0.3477 mol
According to the stoichiometry of the balanced equation, the molar ratio between Sb2S3 and Sb4O6 is 3:2. Therefore, the moles of Sb4O6 that would be formed if the reaction went to completion would be:
Moles of Sb4O6 = (Moles of Sb2S3) × (2/3)
Moles of Sb4O6 = 0.3477 mol × (2/3) ≈ 0.2318 mol
Now, let's convert the moles of Sb4O6 to grams:
Mass of Sb4O6 = Moles of Sb4O6 × Molar mass of Sb4O6
Mass of Sb4O6 = 0.2318 mol × 747.06 g/mol ≈ 173.24 g
So, the theoretical yield of Sb4O6 is approximately 173.24 g.
Finally, to calculate the percent yield, we divide the actual yield (74 g) by the theoretical yield (173.24 g) and multiply by 100:
Percent yield = (Actual yield / Theoretical yield) × 100
Percent yield = (74 g / 173.24 g) × 100 ≈ 42.7%
Therefore, the percent yield for the reaction is approximately 42.7%.