N _ 2(g) + 3 H _ 2(g) ⇌ 2 NH _ 3(g) 20g5g Consider the above...

${N}_{2\left(g\right)} + 3{H}_{2\left(g\right)} ⇌ 2{NH}_{3\left(g\right)}20g5gGivenmass\dfrac{20}{28}\dfrac{5}{2}Numberofmoles$

To find the limiting reagent we can divide given number of moles with Stoichiometric coefficients, whichever will be the lower will be limiting reagent.

${N}_{2} \to \dfrac{\dfrac{20}{28}}{1} = \dfrac{20}{28}{H}_{2} \to \dfrac{\dfrac{5}{2}}{3} = \dfrac{5}{6}$

$∴ {N}_{2}$ is the Limiting Reagent.

According to the reaction one mole of nitrogen gas gives two moles of ammonia.

$∴ n\left({NH}_{3}\right) = 2 \times n\left({N}_{2}\right) = 2 \times \dfrac{20}{28}$

Number of moles of ammonia formed will be $= 1 . 42$