Suppose hydrochloric acid reacts with potassium sulfite yielding water, sulfur dioxide, and potassium chloride. Suppose 4 moles
1 answer:
Answer:
128 grams of sulfur dioxide are produced.
Explanation:
Moles of HCl = 4 moles
According to reaction, 2 moles of HCl gives 1 mole of sulfur dioxide gas.
Then 4 moles of HCl will give:
of sulfur dioxide gas.
Mass of sulfur dioxide gas = 2 mol × 64 g/mol = 128 g
128 grams of sulfur dioxide are produced.
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Question is incomplete, complete question is as follows :
Complete Question : .Sulfonation of benzene has the following mechanism:
(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3
[fast]
(2) SO3 + C6H6 → H(C6H5+)SO3−
[slow]
(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4
[fast]
(4) C6H5SO3− + H3O+ → C6H5SO3H + H2O
[fast]
write the overall rate law for the initial rate of the reaction as a fraction.
Rate=k(________/_________)
Answer:
The overall rate law for the initial reaction is =
Explanation :
Frist of all, all the common terms are cancelled out and written the overall reaction.
As we know that the rate depednant step is the slowest step of the reaction, rate law is :
rate =
But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.
Rate law for synthesis of S03 is as follows :
rate =
Hence when we substitute equation 2 in equation one,
Rate comes out to be =