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8.0 dm of NO is mixed with 6.0 dm3 of O, at room temperature and pressure (RTP).

The reaction below takes place until one of the reactants is used up.

2NO(g) + O2(g) + 2NO,(9)

What is the volume of the mixture at RTP after the reaction has taken place?


A)8.0 dm3

B)10.0 dm3

C)12.0 dm3

D)14.0 dm3

Sagot :

Pstnonsonjokuidn

Answer:

B)10.0 dm3

Explanation:

Now we have to apply Gay-Lussac's law which states that gases combine in simple volumes provided the temperature and pressure remain constant.

                                               2NO(g) + O2(g) ------>2NO

combining volumes                2            :   1               :  2

Volumes before reaction       8.0            6.0               0.0

Reacting volumes                   8.0            4.0                8.0

Volumes after reaction           -                 2.0              8.0

Therefore;

Total volume after reaction = 2.0 + 8.0 = 10.0 dm^3

Eduard22slyidn

The volume of the mixture after the reaction is  10 dm³

The correct answer to the question is Option B. 10 dm³

The above is simply a demonstration of Gay-Lussac' law. The volume of the mixture after the reaction can be obtained as follow:

                            2NO(g) + O₂ —> 2NO(g)

Combining Vol:   2 dm³  :  1 dm³   :  2 dm³

Before reaction:  8 dm³  :  6 dm³  :  0

During reaction:  8 dm³  :  4 dm³  :  8 dm³

After reaction:    0 dm³  :  2 dm³ :  8 dm³

The Volume after the reaction = 2 + 8 = 10 dm³

Learn more about Gay-Lussac's law:

https://brainly.com/question/16984126

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