Consider the following balanced reaction between hydrogen and nitrogen to form ammonia:

3H2(g)+N2(g)→2NH3(g)

How many moles of NH3 can be produced from 21.0 mol of H2 and excess N2?

14 moles of NH3 can be produced from 21.0 mol of H2 and excess N2.

Explanation:

Balanced equation for the production of ammonia is given as:

3H2(g)+N2(g)→2NH3(g)

Given that:

number of moles of H2 = 21 moles

moles of NH3 produced from 21 moles of H2 =?

In the above equation the limiting reagent is hydrogen gas so the production of ammonia directly depends on it.

From the chemical equation we can see that:

3 moles of H2 react to form 2 moles of NH3

so, 21 moles of  H2 will react to form x moles of NH3

=

=

= 14 MOLES

from the calculation by applying stoichiometry it is found that 14 moles of ammonia is produced by using 21 moles of hydrogen.

the answer is the first option

explanation:

just googled it m8. not that hard

lol