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.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.