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

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:

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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

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