Question

56 g of nitrogen and 20 g of hydrogen
reacts in a closed vessel to form
ammonia. Maximum amount of ammonia
produced in this reaction is
17 g
34 g
68 g
51 g​

Answer 1

Answer:

34g

Explanation:

this is the answer...mark as brainlist

Answer 2

Answer:

N2 + 3H₂ → 2NH₃

56g of N2 and 20 g of hydrogen present at initially.

1. find the moles

      moles of N₂ =  mass present /  molar mass =  56/28 = 2moles

      moles of H₂= mass present  /  molar mass  =   20 / 2 =  10moles

limiting reagent is N₂  from above we can conclude that.

now, for 1 mole of N₂ consumed , 2 moles of  NH₃ is produced

so, for 2 moles of N₂  consumed 2 x 2 =  4 moles

now for mass of NH₃ will be =  moles X molar mass

                                              =  4 X 17

                                              =   68g

so option  c is correct answer.