N2g+3H2g=2NH3g , what is the maximum mass that can be produced from a mixture of 2.00×10'3g ,N2 and 8.00×10'4g ,H2

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g)

moles of N₂ used = 2.00x10³ g x 1 mole/28 g = 71.4 moles N₂

moles H₂ used = 8.00x10⁴ g x 1 mole/2 g = 4x10⁴ moles H₂

Limiting reactant is N₂ since it will run out first based on the 3:1 mole ratio of H₂ to N₂ required.

maximum mass of NH₃ produced = 71.4 moles N₂ x 2 moles NH₃/mole N₂ x 17.0 g/mole NH₃ = 2428 g NH₃

Answer would be 2430 g or 2.43x10³ g (to 3 significant figures)