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  • <img alt="\mathrm{ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} }" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%2

\mathrm{ \mathrm{N}_{2(\mathrm{~g})}+3 \mathrm{H}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NH}_{3(\mathrm{~g})} }
20 \mathrm{~g} \quad 5 \mathrm{~g}
Consider the above reaction, the limiting reagent of the reaction and number of moles \mathrm{\mathrm{NH}_{3}} formed respectivly are :

Option: 1

\mathrm{\mathrm{H}_{2}, 1.42 \text { moles }}


Option: 2

\mathrm{\mathrm{H}_{2}, 0.71 \text { moles }}


Option: 3

\mathrm{\mathrm{N}_{2}, 1.42 \text { moles }}


Option: 4

\mathrm{\mathrm{N}_{2}, 0.71 \text { moles }}


Answers (1)

best_answer

\mathrm{N_{2}(g)+3H_{2}(g)\rightleftharpoons 2 NH_{3}(g)}

\mathrm{20 g}               \mathrm{5 g}
\downarrow                     \downarrow
\mathrm{\frac{20}{28}mole}        \mathrm{\frac{5}{6}mole}
\downarrow                     \downarrow
\mathrm{0.71mole}      \mathrm{0.83mole}

\mathrm{0.71< 0.83\: Hence\: N_{2}} is Limiting reagent at this reaction.

Product depend on limiting reagent.
From 1 ,ol of \mathrm{ N_{2}} , 2 mole of \mathrm{ NH_{3}} is forming.
So, from 0.71 mol \mathrm{ N_{2}} , 1.42 mole of \mathrm{ NH_{3}} will form. 
         
So, option (3) is correct.

Posted by

Rakesh

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