Question

How many grams of NH3 can be produced from 4.45mol of N2 and excess H2.

Answer 1

first write REACTION ,
\[\large \bf N_2+3H_2\rightarrow 2NH_3\]
then,find LIMITING REAGENT.
given,H2 is in excess ,so N2 is limiting.

so find moles of NH3 with the help of LIMITING REAGENT i.e N2
\[\large \bf 1~mol~N_2 \rightarrow 2~mole~NH_3\]
so,
\[\large \bf 4.45~mol~N_2\rightarrow 4.45 \times 2~mol~NH_3=8.9~mol~NH_3\]

now,we have moles of NH3 and molecular weight of NH3,now find weight of substance.

\[\large \bf mole=\frac{weight}{molecular~weight}\]
\[\large \bf weight=mole \times molecular~weight=8.9 \times 17=151.3\]
so,
\[\large \bf 151.3g~of ~NH_3~is~produced=\color{red}{Answer}\]

Answer 2

hope you understand. @ipelayo1995

Answer 3

Theoretically correct anwer.

Synthesis of ammonia is an equilibrium reaction,
so yield is not necessarily 100 %.