What amount of hydrogen gas would be necessary to to produce 14.0 mol NH3(g)(assuming there is enough nitrogen gas to complete the reaction)?

Amount of hydrogen gas needed 42 g.

Explanation:

Given data:

Amount of ammonia produced = 14.0 mol

Amount of hydrogen gas needed = ?

Amount of nitrogen = excess

Solution:

Chemical equation:

N₂ + 3H₂      →    2NH₃

Now we will compare the moles of hydrogen with ammonia.

            NH₃              :             H₂

              2                :               3

             14.0            :            3/2×14.0 = 21 mol

Mass of hydrogen needed:

Mass = number of moles × molar mass

Mass = 21 mol × 2 g/mol

Mass = 42 g

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

the temperature of the gas, in k, not celsius.