Answer:
5.42g, 71.77%
Explanation:
First, we have to write out the balanced chemical equation. The unbalanced equation can be written as “SO2+O2 -> SO3” and to balance it, we can see that having two mols of SO2 and two mols of SO3 will make each side have the same amount of mols per element on each side. So the balanced chemical equation is “2SO2 + O2 -> 2SO3”
Now, we want to solve for the theoretical yield in grams of SO3. To do this, we have to use dimensional analysis. We convert g SO2 into mols SO2 using the molar mass of the elements. Then we convert mols of SO2 into mols of SO3 using the balanced equation. Once we’ve done that, we can convert mols of SO3 into grams of SO3.
You should know how to look up the molar mass of elements on the periodic table by now. Find the masses and set up the terms so they cancel like so:
Doing the math, we get 5.42g so3 as the theoretical yield. This is the most amount that you could ever get if the world was a perfect place. But alas, it isn’t and mistakes are gonna happen, so the number is going to be less than that. So the best we can do, is to figure out the percent yield that we got.
In a lab scenario, this was calculated to be 3.89 g as stated by the problem. The percent composition formula is
and plugging the numbers into it, we get:
make sure to follow the decimal/significant figure rules of your instructor, but only round at the end. My professor didn't care too much thankfully, but some professors do
Don’t press the link above me. It causes viruses
Answer:
V2 = 3.11 x 105 liters
Explanation:
Initial Volume, V1 = 2.16 x 105 liters
Initial Temperature, T1 = 295 K
Final Temperature, T2 = 425 K
Final Volume, V2 = ?
These quantities are related by charle's law and the equation of the law is given as;
V1 / T1 = V2 / T2
V2 = T2 * V1 / T1
V2 = 425 * 2.16 x 105 / 295
V2 = 3.11 x 105 liters
Answer:
B. First order, Order with respect to C = 1
Explanation:
The given kinetic data is as follows:
A + B + C → Products
[A]₀ [B]₀ [C]₀ Initial Rate (10⁻³ M/s)
1. 0.4 0.4 0.2 160
2. 0.2 0.4 0.4 80
3. 0.6 0.1 0.2 15
4. 0.2 0.1 0.2 5
5. 0.2 0.2 0.4 20
The rate of the above reaction is given as:
![Rate = k[A]^{x}[B]^{y}[C]^{z}](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BA%5D%5E%7Bx%7D%5BB%5D%5E%7By%7D%5BC%5D%5E%7Bz%7D)
where x, y and z are the order with respect to A, B and C respectively.
k = rate constant
[A], [B], [C] are the concentrations
In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.
Order w.r.t A : Use trials 3 and 4
![\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}](https://tex.z-dn.net/?f=%5Cfrac%7BRate3%7D%7BRate4%7D%3D%20%5B%5Cfrac%7B%5BA%283%29%5D%7D%7B%5BA%284%29%5D%7D%5D%5E%7Bx%7D)
![\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}](https://tex.z-dn.net/?f=%5Cfrac%7B15%7D%7B5%7D%3D%20%5B%5Cfrac%7B%5B0.6%5D%7D%7B%5B0.2%5D%7D%5D%5E%7Bx%7D)
Order w.r.t B : Use trials 2 and 5
![\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}](https://tex.z-dn.net/?f=%5Cfrac%7BRate2%7D%7BRate5%7D%3D%20%5B%5Cfrac%7B%5BB%282%29%5D%7D%7B%5BB%285%29%5D%7D%5D%5E%7By%7D)
![\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}](https://tex.z-dn.net/?f=%5Cfrac%7B80%7D%7B20%7D%3D%20%5B%5Cfrac%7B%5B0.4%5D%7D%7B%5B0.2%5D%7D%5D%5E%7By%7D)
Order w.r.t C : Use trials 1 and 2
![\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}](https://tex.z-dn.net/?f=%5Cfrac%7BRate1%7D%7BRate2%7D%3D%20%5B%5Cfrac%7B%5BA%281%29%5D%7D%7B%5BA%282%29%5D%7D%5D%5E%7Bx%7D%5B%5Cfrac%7B%5BB%281%29%5D%7D%7B%5BB%282%29%5D%7D%5D%5E%7By%7D%5B%5Cfrac%7B%5BC%281%29%5D%7D%7B%5BC%282%29%5D%7D%5D%5E%7Bz%7D)
we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:
![\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}](https://tex.z-dn.net/?f=%5Cfrac%7B160%7D%7B80%7D%3D%20%5B%5Cfrac%7B%5B0.4%5D%7D%7B%5B0.2%5D%7D%5D%5E%7B1%7D%5B%5Cfrac%7B%5B0.4%5D%7D%7B%5B0.4%5D%7D%5D%5E%7B2%7D%5B%5Cfrac%7B%5B0.2%5D%7D%7B%5B0.4%5D%7D%5D%5E%7Bz%7D)
z = 1
Therefore, order w.r.t C = 1
Hey there!
Exothermic reactions release heat, causing its temperature to fall. If the reaction is lessening the temperature of the object while releasing all the heat, then you know that the reaction is indeed exothermic. Hope this helps!
