The value of Keq for this reaction is ![\bold{1.8 \times 10^-^5}](https://tex.z-dn.net/?f=%5Cbold%7B1.8%20%5Ctimes%2010%5E-%5E5%7D)
The correct option is (A).
<h3>What is acetic acid?</h3>
Acetic acid is an organic compound also called ethanoic acid. It is a colorless liquid with a chemical formula
.
It has antibacterial and antifungal properties.
When acetic acid is reacted with water, it completely dissolves in it.
Thus, the correct option is A,
.
Learn more about acetic acid, here:
brainly.com/question/12924347
Answer:
It is referred to as b) premises liability
Sex
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Balanced:
2NaOH + CO2 → Na₂CO₃ + H₂O
Explanation:
To balance an equation, the amount of atoms on each side must be <u>equal.</u>
Atom Count on Reactant Side:
Atom Count On Product Side:
As you see, to simply balance the equation, put the 2 in order for the atom count on the reactant side to be equal to the product side. The 2 will add one more of each atom so that its the same as the product side.
The given question is incomplete. The complete question is :
Gaseous butane reacts with gaseous oxygen gas to produce gaseous carbon dioxide and gaseous water . If 1.31g of water is produced from the reaction of 4.65g of butane and 10.8g of oxygen gas, calculate the percent yield of water. Be sure your answer has the correct number of significant digits in it.
Answer: 28.0 %
Explanation:
To calculate the moles :
![\text{Moles of butane}=\frac{4.65g}{58g/mol}=0.080moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20butane%7D%3D%5Cfrac%7B4.65g%7D%7B58g%2Fmol%7D%3D0.080moles)
![\text{Moles of oxygen}=\frac{10.8g}{32g/mol}=0.34moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20oxygen%7D%3D%5Cfrac%7B10.8g%7D%7B32g%2Fmol%7D%3D0.34moles)
According to stoichiometry :
13 moles of
require 2 moles of butane
Thus 0.34 moles of
will require=
of butane
Thus
is the limiting reagent as it limits the formation of product and butane is the excess reagent.
As 13 moles of
give = 10 moles of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Thus 0.34 moles of
give =
of ![H_2O](https://tex.z-dn.net/?f=H_2O)
Mass of ![H_2O=moles\times {\text {Molar mass}}=0.26moles\times 18g/mol=4.68g](https://tex.z-dn.net/?f=H_2O%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.26moles%5Ctimes%2018g%2Fmol%3D4.68g)
![{\text {percentage yield}}=\frac{\text {Experimental yield}}{\text {Theoretical yield}}\times 100\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B%5Ctext%20%7BExperimental%20yield%7D%7D%7B%5Ctext%20%7BTheoretical%20yield%7D%7D%5Ctimes%20100%5C%25)
![{\text {percentage yield}}=\frac{1.31g}{4.68g}\times 100\%=28.0\%](https://tex.z-dn.net/?f=%7B%5Ctext%20%7Bpercentage%20yield%7D%7D%3D%5Cfrac%7B1.31g%7D%7B4.68g%7D%5Ctimes%20100%5C%25%3D28.0%5C%25)
The percent yield of water is 28.0 %