Answer:
Explanation:
Hello!
In this case, given the chemical reaction:
In such a way, given the volumes and molarities of each reactant, we can compute the moles of produced iron (III) hydroxide by each of them, via the 3:1 and 1:1 mole ratios:
It means that the sodium hydroxide is the limiting reactant and 0.00833 moles of iron (III) hydroxide are produced; thus, the required mass is:
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.
Learn more about percent yield at: brainly.com/question/8638404
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