How many grams of nh3 can be produced from 3.72 mol of n2 and excess h2.

1 answer:

7 0

N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol

n = 7,44mol
M = 17g/mol

m = n * M = 7,44mol * 17g/mol = 126,48g

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