contestada

A sample of gas contains 0.1000 mol of N2(g) and 0.3000 mol of H2(g) and occupies a volume of 17.4 L. The following reaction takes place: N2(g) + 3H2(g)2NH3(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

Respuesta :

Answer:

V2=3.482L

Explanation:

The reaction takes place as follows

            [tex]N2 + 3H2[/tex]    →  [tex]2NH3[/tex]

In order to find the volume of this sample after the reaction we will use the

Avogadro's law

                    [tex]\frac{V1}{n1}=\frac{V2}{n2}[/tex]

V1=17.4L

n1= moles of N2 +3H2= 0.1+3*0.3=1mole

V2= to be found

Now to calculate the n2 of NH3 we will use the reaction  

as we know that N2 is limiting reagent and hydrogen is present in excess so the ammonia will only be formed by this limiting nitrogen

1mole of N2 gives 2 mole of NH3

0.1*2= 0.2mole of NH3 will be produced

n2 = 0.2

upon substituting theses values in the Avogadro law

[tex]\frac{17.4}{1} =\frac{V2}{0.2}[/tex]

V2=3.482L

 

Otras preguntas