Question

7.Calculate the volume of nitrogen that reacts with 12dm3 of hydrogen with the volume of both gases measured at rtp:

Answer 1

Answer:

4.03dm³

Explanation:

The reaction expression is given as:

       3H₂   +   N₂   →    2NH₃  

  Volume of hydrogen  = 12dm³  

AT rtp:

             1 mole of gas occupies volume of 22.4dm³  

             x mole of hydrogen will occupy a volume of 12dm³

     Number of moles of hydrogen  = $\frac{12}{22.4}$   = 0.54mole

From the balanced reaction equation:

            3 mole of hydrogen gas combines with 1 mole of Nitrogen gas

         0.54 mole of hydrogen as will therefore combine with $\frac{0.54}{3}$   = 0.18moles of nitrogen gas

Since ;

                     1 mole of gas occupies a volume of 22.4dm³

               0.18moles of Nitrogen gas will occupy 0.18 x 22.4  = 4.03dm³