N2(g)+3H2(g)⇌2NH3(g)20g5gGivenmass282025Numberofmoles
To find the limiting reagent we can divide given number of moles with Stoichiometric coefficients, whichever will be the lower will be limiting reagent.
N2→12820=2820H2→325=65
∴N2 is the Limiting Reagent.
According to the reaction one mole of nitrogen gas gives two moles of ammonia.
∴n(NH3)=2×n(N2)=2×2820
Number of moles of ammonia formed will be =1.42