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  • How to solve this problem- 28 g N2and 6 g H2were mixed. At equilibrium 17 g NH3was formed. The mass of N2and H2 of equilibrium are

28 g N2 and 6 g H2 were mixed. At equilibrium 17 g NH3 was formed. The mass of N2 and H2 of equilibrium are 

  • Option 1)

    11 g , zero

  • Option 2)

    1 g , 3 g

  • Option 3)

    14 g , 3 g

  • Option 4)

    11 g , 3 g

 

Answers (1)

best_answer

As we learned in concept

Law of Chemical equilibrium -

At a given temperature, the product of concentration of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentration of the reactants raised to their individual stoichiometric coefficients has a constant value.

- wherein

aA+bB\rightleftharpoons cC+dD


K_{c}=\frac{[C]^{c\:[D]^{d}}}{[A]^{a}\:[B]^{b}}

[A],\:[B],\:[C]\:[D]

are equilibrium concentration

 

 The reaction goes as follows:

                           N_{2}+3\textrm{H}_{2}\:=\:2NH_{3}

t = 0                    1             3                0

At equilibrium      1/2         3/2               1

\therefore \:Mass\:of\:N_{2}\:at\:equilibrium\:=\:28\times \frac{1}{2}\:=\:14\:g

\therefore \:Mass\:of\:H_{2}\:at\:equilibrium\:=\:2\times \frac{3}{2}\:=\:3\:g


Option 1)

11 g , zero

This option is incorrect.

Option 2)

1 g , 3 g

This option is incorrect.

Option 3)

14 g , 3 g

This option is correct.

Option 4)

11 g , 3 g

This option is incorrect.

Posted by

prateek

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