Question

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

3H2 + N2 -----> 2NH3

Answer 1

4.03dm³

Step-by-step 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 =
`(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
`(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