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:
True
Explanation:
Because atoms with the same number of valence electrons react in similar ways with other elements.
Explanation:
Is the loss of electrons, gains of oxygen or loss of hydrogen
Answer: The rate law is ![rate=k[CH_3COOC_2H_5]^1[NaOH]^1](https://tex.z-dn.net/?f=rate%3Dk%5BCH_3COOC_2H_5%5D%5E1%5BNaOH%5D%5E1)
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
For the given reaction:
k= rate constant
Rate law: ![rate=k[CH_3COOC_2H_5]^x[NaOH]^y](https://tex.z-dn.net/?f=rate%3Dk%5BCH_3COOC_2H_5%5D%5Ex%5BNaOH%5D%5Ey)
For the given rate law:
y =1 = order with respect to 
n = total order = 2
2= (x+y)
2= (x+1)
x= 1
Thus order with respect to 
is 1 and rate law is :
