For a reaction,

N_{2}(g)+3H_{2}(g)\rightarrow 2NH_{3}(g);\, \, \, \, \, \: \: \; \: \:

identify dihydrogen \left ( H_{2} \right ) as a limiting reagent in the following reaction mixtures.

  • Option 1)

     56\, g\: of\: N_{2}\: +\: 10\, g\: of\: H_{2}              

  • Option 2)

    35\: g\: of\: N_{2}\: +\: 8\: g\: of\: H_{2}

  • Option 3)

    28\: g\: of\: N_{2}\: +\: 6\: g\: of\: H_{2}

  • Option 4)

    14\: g\: of\: N_{2}\: +\: 4\: g\: of\: H_{2}

Answers (1)

 

Concept of limiting reagent and excess reagent -

The reactant which gets consumed and thus limits the amount of product formed is called the limiting reagent.

- wherein

 CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g) ;

In this reaction, if 1 mole of methane and 1 mole of oxygen are taken then oxygen would be limiting reagent.

 

 

 

For :-

N_{2}\left ( g \right )+3H_{2}\left ( g \right )\rightarrow 2NH_{3}\left ( g \right );

Identifying H_{2} as a limiting reagent

(1)56\: g\: of\: N_{2}\: +\: 10\: g\: of\: H_{2}

       \frac{56}{28}=2\; mole \; \; \; \; \frac{10}{2}=5\; mole

(2)35\: g\: of\: N_{2}\: +\: 8\: g\: of\: H_{2}

      \frac{35}{28}=1.25\; mole \; \; \; \; \frac{8}{2}=4\; mole

       (LR)

(3)28\: g\: of\: N_{2}\: +\: 6\: g\: of\: H_{2}      :- Reaction gets completed here

      \frac{28}{28}=1\; mole \; \; \; \; \frac{6}{2}=3\; mole

(4) 14\: g\: of\: N_{2}\: +\: 4\: g\: of\: H_{2}

     \frac{14}{28}=0.5\; mole \; \; \; \; \frac{4}{2}=2\; mole

      (LR)

 

 


Option 1)

 56\, g\: of\: N_{2}\: +\: 10\, g\: of\: H_{2}              

Option 2)

35\: g\: of\: N_{2}\: +\: 8\: g\: of\: H_{2}

Option 3)

28\: g\: of\: N_{2}\: +\: 6\: g\: of\: H_{2}

Option 4)

14\: g\: of\: N_{2}\: +\: 4\: g\: of\: H_{2}

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