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.
56g of N2 reacts with 30g of H2 to form NH3.Calculate the number of moles of NH3 formed.
1 answer
1 answer