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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<u>Answer:</u>
(A)
Density = Mass / Volume
So
Mass = Density × Volume


Mole ratio of toluene : Oxygen is 1 : 9

(B)
1 mole of Toluene produces 7 moles of
gas and 4 moles of
Vapour
So the mole ratio is 1 : 11

(C)
1mole contains
molecules

Answer:
Yes
Explanation:
Natural gas is colorless and odorless, and explosive, so a sulfur-smell (similar to rotten eggs) is usually added for early detection of leaks. ... Natural gas is a fossil fuel. Natural gas is a non-renewable hydrocarbon used as a source of energy for heating, cooking, and electricity generation.
Adding (S2O3)2- would affect the reaction mechanism that involves this ion. From the reaction mechanism given above, the equilibrium of step 2 would be affected. Adding the stock solution of (S2O3)2- would shift the equilibrium to the right thus making more products of the said mechanism. Also, the reaction rate of this step would occur faster than the original rate. This is based on Le Chatelier's Prinicple which states that a corresponding change would happen to the equilibrium of a reaction when pressure, concentration of the substances or temperature is changed. So, that after the addition, a color change would appear immediately because I3- would be removed slowly from solution, and would therefore be able to react with starch.
<h3><u>Answer;</u></h3>
= 12.5 Moles of CaSO3
<h3><u>Explanation</u>;</h3>
The reaction between CaCO3 and SO2 is given by the equation.
CaCO3(s) + SO2(g) → CaSO3(aq) + CO2(g)
The mole ratio between CaCO3 and SO2 is 1 : 1;
1 mole of CaCO3 reacts with 1 mole SO2 to form CaSO3 and CO2
Therefore;
<em>12.5 moles of SO2 will require 12.5 moles of CaSO3</em>