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,

N_{2} + 3H_{2 }---------> 2NH_{3}

Number of moles of N_{2} = 50kg X 1000g/1kgx1mole/28g

= 17.86 x 10^{2} mol

Number of moles of H_{2 }= 10kg x1000g/1kgx 1mol/2.016g

= 4.96 x 10^{3} mol

According to the balanced reaction :

17.86 x 10^{2} mol of N_{2} will react with = 3/1 x 17.86 x 10^{2}

= 5.36 x 10^{3 }mol

since number of moles of H_{2} are 4.96 x 10^{3}

Therefore dihydrogen is the limiting reagent.

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