Asked by Anina Anna Charley
56g of N2 reacts with 30g of H2 to form NH3.Calculate the number of moles of NH3 formed.
Answers
Answered by
DrBob222
N2 + 3H2 ==> 2NH3
mols N2 to start = g/molar mass = 56/17 = 3.3
mols NH3 formed if N2 is the limiting reagent = 3.3 x (2 mols NH3/1 mol N2) = 3.3 x 2 = 6.6
mols H2 to start = 30/2 = 15.
mols NH3 formed if H2 is the limiting reagent = 15 x (2 mols NH3/3 mols H2) = 10.
So N2 is the limiting reagent and 6.6 mols NH3 will be formed. Some H2 will be unreacted at the end.
mols N2 to start = g/molar mass = 56/17 = 3.3
mols NH3 formed if N2 is the limiting reagent = 3.3 x (2 mols NH3/1 mol N2) = 3.3 x 2 = 6.6
mols H2 to start = 30/2 = 15.
mols NH3 formed if H2 is the limiting reagent = 15 x (2 mols NH3/3 mols H2) = 10.
So N2 is the limiting reagent and 6.6 mols NH3 will be formed. Some H2 will be unreacted at the end.