How many grams of NH3 can be produced from 4.46 mol of N2 and excess H2.?

Answer 1

The molecular weight of NH3 is 17.03-grams per mole and 14.01 for N2. The reaction is N2 + 3H2 = NH3. Therefore for every 1-mole of N2 as a reactant 1-mole of NH3 is produced. .2941-moles of NH3 is produced with a mass of 5.01-grams.

Answer 2

1 mol of N2 reacts with 3 mol of H2 to produce 2 mol of NH3. Therefore, 4.46 mol of N2 will produce (4.46 mol x 2 mol NH3 / 1 mol N2) = 8.92 mol of NH3. As NH3 has a molar mass of 17.03 g/mol, 8.92 mol of NH3 is equal to (8.92 mol x 17.03 g/mol) = 152 g of NH3.

Answer 3

Balanced equation: 3H2 + N2 ==> 2NH34.16 moles N2 x 2 mole NH3/1 mole N2 = 8.32 moles NH3

8.32 moles NH3 x 17 g/mole = 141 grams NH3

Answer 4

9,62 moles ammonia = 163.84 g

Answer 5

The answer is 157,37 g ammonia.

Answer 6

151,9 g ammonia

Answer 7

143

Answer 8

148g NH3