How many grams of NH3 can be produced from 4.01 mol of N2 and excess H2.

Question

Question

Answer 1

Answer #1

the reaction

N2 + 3H2 = 2NH3

mol of N2 --> 4.01

ratio is 1:2

so

mol of NH3 = 2*N2 = 2*4.01 = 8.02 mol of NH3

mass = mol*MW = 8.02*17.031 = 136.5886 g of NH3 can be produced

Answer 2

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