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tatiyna
3 years ago
14

potassium chlorate (kclo3) decomposes into potassium chloride (kcl) and oxygen gas (o2) how many grams of oxygen can be produced

from the decomposition of 7.38 moles of potassium chlorate​
Chemistry
2 answers:
sergeinik [125]3 years ago
7 0

Answer:

m_{O_2}=354.24gO_2

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)

In such a way, as 7.38 moles of potassium chloride are decomposed, the resulting grams of oxygen are computed considering a 2 to 3 molar relationship in the chemical reaction:

m_{O_2}=7.38molKClO_3*\frac{3molO_2}{2molKClO_3}*\frac{32gO_2}{1molO_2} \\\\m_{O_2}=354.24gO_2

Best regards.

gladu [14]3 years ago
5 0

Consider this reaction  : 2KClO3 → 2KCl + 3O2  

<em>Mole ratio</em> = 3(O2) : 2(KClO3)

<em>Number of moles of O2</em> = 3/2*7.38 = 11.07 mol

<em>Number of moles of 02</em> = mass/molar mass

<em>Therefore, mass</em> = 11.07*(16*2) = 354.24 g

Note: It's mandatory to always balance the reaction at first

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What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

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Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

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