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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Answer:
3.89 kg P2O5 must be used to supply 1.69 kg Phosphorus to the soil.
Explanation:
The molecular mass of P2O5 is
P2 = 2* 31 = 62
O5 = 5 *<u> 16 = 80</u>
Molecular Mass = 142
Set up a Proportion
142 grams P2O5 supplies 62 grams of phosphorus
x kg P2O5 supplies 1.69 kg of phosphorus
Though this might be a bit anti intuitive, you don't have to convert the units for this question. The ratio is all that is important.
142/x = 62/1.69 Cross multiply
142 * 1.69 = 62x combine the left
239.98 = 62x Divide by 62
239.98/62 = x
3.89 kg of P2O5 must be used.
