CBSE Class 11-science Answered
50.0kg of N2[g] and 10.0 kg of H2 [g] are mixed to produce NH3[g] formed.Identify the limiting reagent in production of NH3 in this situaton
Asked by virubloda6 | 21 May, 2019, 08:39: AM
Expert Answer
Given:
Mass of N2 = 50 kg
= 50 × 103 gm
No. of moles of N2
Mass of H2 = 10 kg
= 10 × 103 gm
No. of moles of H2
Formation of ammonia;
From reaction, 1 mole of N2 requires 3 moles H2.
So 1.78× 103 mole of N2 will require (1.78×103 ×3) = 5.34 ×103 moles of H2.
Here 5×103 moles of H2 are given, therefore H2 is the limiting reagent.
Now 5×103 moles of H2 will give,
Therfore the amount of NH3 produced is 3.33×103 mole.
Answered by Varsha | 21 May, 2019, 11:52: AM
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