How many grams of both nitrogen and hydrogen gas are required to produce 400.0 grams of ammonia gas

Answer 1

First, look at the balanced equation for this synthesis reaction.

N₂(g) + 3H₂(g) ==> 2NH₃(g)

Convert mass of NH₃ to moles, and then determine moles of N₂ and H₂ needed.

400.0 g NH₃ x 1 mole NH₃/17.031 g = 23.487 moles NH₃ gas

moles N₂ needed = 23.487 moles NH₃ x 1 mole N₂/2 moles NH₃ = 11.7435 moles N₂ needed

moles H₂ needed = 23.487 moles NH₃ x 3 moles H₂/2moles NH₃ = 35.2305 moles H₂ needed

Finally, convert moles of N₂ and H₂ to grams and correct for significant figures.

N₂ needed = 11.7435 moles N₂ x 14.0067 g/mole = 164.5 g N₂ needed (to 4 significant figures)

H₂ needed = 35.2305 moles H₂ x 2.0159 g/mole = 71.02 g H₂ needed (to 4 significant figures)