The electrolysis of molten alcl3 for 3.25 hr with an electrical current of 15.0 a produces ________ g of aluminum metal.
1 answer:
Based on Faraday's 1st law of electrolysis,
total electric current passed in AlCl3 = I = Q X T
where Q = electric charge, T = time (in seconds)
Given, Q = 15 A, T = 3.5 h = 3.5 X 60 X 60 s = 12600 s
∴ I = 15 X 12600 = 189000 C
Now, based on Faraday's 2nd law of electrolysis we have,
W = ZIT
where, W= weight of electrolyte deposited/evolved at any electrode
Z= electrochemical equivalent
T= time (s)
I = electric current
in present case, Electrochemical Equivalent of Al =
= 8.993 g and I = 189000 C
∴ W = 8.993 X 189000 X 12600 = 2.141X
g
total electric current passed in AlCl3 = I = Q X T
where Q = electric charge, T = time (in seconds)
Given, Q = 15 A, T = 3.5 h = 3.5 X 60 X 60 s = 12600 s
∴ I = 15 X 12600 = 189000 C
Now, based on Faraday's 2nd law of electrolysis we have,
W = ZIT
where, W= weight of electrolyte deposited/evolved at any electrode
Z= electrochemical equivalent
T= time (s)
I = electric current
in present case, Electrochemical Equivalent of Al =
∴ W = 8.993 X 189000 X 12600 = 2.141X
You might be interested in
Answer:
49.2 g/mol
Explanation:
Let's first take account of what we have and convert them into the correct units.
Volume= 236 mL x () = .236 L
Pressure= 740 mm Hg x ()= 0.97 atm
Temperature= 22C + 273= 295 K
mass= 0.443 g
Molar mass is in grams per mole, or MM= or MM=
. They're all the same.
We have mass (0.443 g) we just need moles. We can find moles with the ideal gas constant PV=nRT. We want to solve for n, so we'll rearrange it to be
n=, where R (constant)= 0.082 L atm mol-1 K-1
Let's plug in what we know.
n=
n= 0.009 mol
Let's look back at MM= and plug in what we know.
MM=
MM= 49.2 g/mol
Answer : The amount of heat released in the combustion is, 209.5 kJ
Explanation :
First we have to calculate the moles of Al and .
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction will be:
From the balanced reaction we conclude that
As, 3 mole of react with 4 mole of
So, 0.188 moles of react with
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and
is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of
From the reaction, we conclude that
As, 3 mole of react to give 2 mole of
So, 0.188 moles of react to give
moles of
Now we have to calculate the amount of heat released in the combustion.
As, 1 mole of releases amount of heat = 1676 kJ
So, 0.125 mole of releases amount of heat =
Thus, the amount of heat released in the combustion is, 209.5 kJ
The structure of alkene is given below in the image.
Alkenes are described as both branched or unbranched hydrocarbons that possess at least one carbon-carbon double bond (CC) and feature a popular system of CnH2n [1].
Alkanes are organic compounds that encompass single-bonded carbon and hydrogen atoms. The formulation for Alkanes is CnH2n+2, subdivided into 3 companies – chain alkanes, cycloalkanes, and branched alkanes.
The geometry around each carbon atom is based totally on a trigonal planar shape, due to the fact, that each carbon has three electrons around it. This must make the angle of each bond one hundred twenty.
Learn more about Alkanes here: brainly.com/question/17040500
#SPJ4
