50.0 kg of N2 (g) and 10.0 kg of H2 (g) are mixed to produce NH3 (g). Calculate the

NH3 (g) formed. Identify the limitingreagent in the production of NH3 in this situation


Dear Student,

N2 + 3H2 ---------> 2NH3

Number of moles of N2 = 50kg X 1000g/1kgx1mole/28g

 = 17.86 x 102 mol

Number of moles of H2 = 10kg x1000g/1kgx 1mol/2.016g

 =  4.96 x 103 mol

According to the balanced reaction :

17.86 x 102 mol of N2 will react with = 3/1 x 17.86 x 102

 = 5.36 x 103 mol

since number of moles of H2 are  4.96 x 103

Therefore dihydrogen is the limiting reagent.

  • -33

the limiting reagent here is h2 coz i is less in quantity !

  • -26

0.5 moles of NH3 ,

i. e . 8.5 gm.  or , 0.0085 kg.

  • -32

actually , i mean to say,

1 mole of N2, combine with 0.8 mole of  H2 to form 0.5 mole of NH3.  and that is 0.0085  kg .

  • -31
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