Assume there is 100 grams of the compound. The amounts of each elements would be: 47.08 g C, 6.59 g H and 46.33 g Cl. Convert them to moles.
mol C: 47.08 g * 1 mol/12 g = 3.9233
mol H: 6.59 g * 1mol/1g = 6.59
mol Cl: 46.33 g * 1mol/35.45 g = 1.3069
Divide the smallest amount of moles to each moles of the elements:
C: 3.9233/1.3069 = 3
H: 6.59/1.3069 = 5
Cl: 1.3069/1.3069 = 1
Therefore, the empirical formula of the compound is C₃H₅Cl.
Given:
Molar mass CO2 is 44.01 g/mol
25.5 g of CO2
Required:
Moles of LiOH
Solution:
Balanced equation is:
2LiOH +CO2 → Li2CO3 + H2O
25.5g CO2 (1 mole CO2/44.01
g/mol CO2)(2 moles LiOH/1 mol CO2) = 1.16 moles LiOH
The equilibrium concentrations of all chemical species at room temperature are given below.
[H2] = 0.212M
[I2] = 0.212M
[HI] = 1.576M
H2(g) + I2(g) <-----> 2HI(g)
I 1 1 0
c -x -x 2x
E 1-x 1-x 2x
[H2] ⇒ 1-x/1 = 1-x
[I2] ⇒ 1-x/1 = 1-x
[HI] ⇒ 2x
Kc = [HI]^2/[H2][I2]
55.3 = (2x)^2/(1-x)(1-x)
55.3 = (2x/1-x)^2
7.44 = 2x/1-x
7.44*(1-x) = 2x
x = 0.788
[H2] ⇒ 1-x = 1-0.788 = 0.212M
[I2] ⇒ 1-x = 1-0.788 = 0.212M
[HI] ⇒ 2x = 2*0.788 = 1.576M
In a chemical reaction, while each reactant and the products are in a concentration that does not alternate with time any extra, it is said to be in a state of chemical equilibrium. in this kingdom, the price of forwarding response is the same as the rate of backward response.
Equilibrium is whilst the fee of the ahead reaction equals the charge of the opposite reaction. All reactant and product concentrations are consistent at equilibrium.
We are saying that a chemical is in an equilibrium concentration while the products and reactants do not change as time moves on. In different words, chemical equilibrium or equilibrium awareness is a state when the price of an ahead reaction in a chemical response will become equal to the rate of a backward response.
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I think it's producers. Not sure, but I would think that something would have to be made before changes can take place.
The total percent yield:
After the combustion reaction with methane, the percent yield was 66.7%.
Combustion of Methane:
- Methane produces a blue flame as it burns in the atmosphere.
- Methane burns in the presence of enough oxygen to produce carbon dioxide (CO₂) and water (H₂O).
- It creates a significant quantity of heat during combustion, making it an excellent fuel source.
The other reactant, air's excess oxygen, is always present, making methane the limiting reactant. As a result, the amount of CH₄ burned will determine how much CO₂ and H₂O are produced.
The following chemical process produces carbon dioxide from methane:
CH₄ + 2O₂ ⇒ CO₂ + 2H₂O
Calculations:
1. <u><em>Theoretical quantity of carbon dioxide:</em></u>
All calculations will be based on the amount of methane because the problem specifies that it is the limiting reagent:
12.0g of CH₄ × (1 mol of CH₄/16g CH₄) × (1 mole of CO₂/1 mole of CH₄) × (44g CO₂/1 mole of CO₂)
= 33g of CO₂
2. <u><em>Percent yield:</em></u>
= Actual yield/Theoretical yield × 100
= 22.0g/33g × 100
= 66.7%
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