detailed solution

Correct option is ඈ

Reaction can be r presented as:  ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$
Moles of nitrogen$=\frac{2.8\times {10}^3}{28}={10}^2\mathrm{moles}$
Moles of hydrogen    $=\frac{1\times {10}^3}{2}=5\times {10}^2\mathrm{moles}$
For ${10}^2$ moles of  ${\mathrm{N}}_2,{\mathrm{H}}_2\mathrm{needed}3\times {10}^2$ moles

 $N_2$ is limiting reagent. 

1 moles of   ${\mathrm{N}}_2$gives 2 moles of NH₃
100  moles of N₂ forms 200 moles of of ${\mathrm{NH}}_3=200\times 17=3400\mathrm{g}$

detailed solution

Correct answer is 3400

Reaction can be r presented as:  ${\mathrm{N}}_2+3{\mathrm{H}}_2\to 2{\mathrm{NH}}_3$
Moles of nitrogen$=\frac{2.8\times {10}^3}{28}={10}^2\mathrm{moles}$
Moles of hydrogen    $=\frac{1\times {10}^3}{2}=5\times {10}^2\mathrm{moles}$
For ${10}^2$ moles of  ${\mathrm{N}}_2,{\mathrm{H}}_2\mathrm{needed}3\times {10}^2$ moles

 $N_2$ is limiting reagent. 

1 moles of   ${\mathrm{N}}_2$gives 2 moles of NH₃
100  moles of N₂ forms 200 moles of of ${\mathrm{NH}}_3=200\times 17=3400\mathrm{g}$