In an experiment, 107.9 grams of H2SO, is produced when 196.2 grams
1 answer:
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The reaction is second order in AB, so: ![v=k[AB]^2](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2)
. In the statement, we obtain that ![[AB]=0.104~M](https://tex.z-dn.net/?f=%5BAB%5D%3D0.104~M)
and, at 25 ºC, 
. Then:
![v=k[AB]^2\\\\ v=0.0164\cdot0.104^2\\\\ v=0.0164\cdot0.010816\\\\ v\approx0.000177=1.77\times10^{-4}~mol/s](https://tex.z-dn.net/?f=v%3Dk%5BAB%5D%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.104%5E2%5C%5C%5C%5C%0Av%3D0.0164%5Ccdot0.010816%5C%5C%5C%5C%0Av%5Capprox0.000177%3D1.77%5Ctimes10%5E%7B-4%7D~mol%2Fs)
Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:
Since each AB molecule forms one of A and one of B,
. Hence: 
.
We'll consider that in the beginning there was not A or B. So,
. Furthermore, since the ratio of AB to A and to B is 1:1, 
.
Calculating the time by the expression of velocity:
![v=\dfrac{|\Delta[AB]|}{\Delta t}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_{AB}|}{V}=\dfrac{1}{\Delta t}\cdot\dfrac{|\Delta n_A||}{250~mL}\\\\ 1.77\cdot10^{-4}=\dfrac{1}{\Delta t}\cdot\dfrac{0.0026~mol}{0.25~L}\\\\ \Delta t=\dfrac{0.0026}{0.25\cdot1.77\cdot10^{-4}}\\\\ \boxed{\Delta t\approx58.76~s}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B%7C%5CDelta%5BAB%5D%7C%7D%7B%5CDelta%20t%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_%7BAB%7D%7C%7D%7BV%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B%7C%5CDelta%20n_A%7C%7C%7D%7B250~mL%7D%5C%5C%5C%5C%0A1.77%5Ccdot10%5E%7B-4%7D%3D%5Cdfrac%7B1%7D%7B%5CDelta%20t%7D%5Ccdot%5Cdfrac%7B0.0026~mol%7D%7B0.25~L%7D%5C%5C%5C%5C%0A%5CDelta%20t%3D%5Cdfrac%7B0.0026%7D%7B0.25%5Ccdot1.77%5Ccdot10%5E%7B-4%7D%7D%5C%5C%5C%5C%0A%5Cboxed%7B%5CDelta%20t%5Capprox58.76~s%7D)
Now, we'll calculate the number of mols of the products in the gas. Using the Ideal Gas Law:
Since each AB molecule forms one of A and one of B,
We'll consider that in the beginning there was not A or B. So,
Calculating the time by the expression of velocity:
Answer:
32.7 grams of Zn will remained in the crucible after cooling.
Explanation:
..[1]
..[2]
Adding [1] + 2 × [2] we get:
..[3]
Moles of ZnS in crucible = 0.50 mol
According to reaction [3]. 2 moles of ZnS gives 2 moles of Zn.
Then 0.50 moles of ZnS will give:
of Zn.
Mass of 0.50 moles of Zn =
= 0.50 mol × 65.4 g/mol =32.7 g
32.7 grams of Zn will remained in the crucible after cooling.