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

Question

mayankdevnani · Answer

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,
$$\large \bf 151.3g~of ~NH_3~is~produced=\color{red}{Answer}$$

mayankdevnani · Answer

hope you understand. @ipelayo1995

gebooors · Answer

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