600 g of dinitrogen reacts with 100 g of dihydrogen to form ammonia. Find the limiting reagent and calculate the amount of ammonia formed? P.S- pls solve the question by calculating the moles

Answer 1

We assess the reaction: #1/2N_2(g) + 3/2H_2(g) rarr NH_3(g)#

#"Moles of dihydrogen"=(100*g)/(2.016*g*mol^-1)=49.6*mol#

#"Moles of dinitrogen"=(600*g)/(28.02*g*mol^-1)=21.4*mol#

It is obvious that there is not enough dihydrogen for a full reduction, and dihydrogen is the reagent with the LIMIT.

One third of an equivalent, i.e. #(49.6*mol)/3=16.5*mol# dinitrogen could react to give ammonia.... Note that we might expect incomplete reduction of dinitrogen in this scenario, i.e. hydrazine, #H_2N-NH_2#, could be quantitatively produced....

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Answer 2

Luna Chen

Dihydrogen is the limiting reagent. The amount of ammonia formed is 300 g.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.