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

1 Answer
anor277
Dec 21, 2017

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

Explanation:

#"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#

Clearly, there is insufficient dihydrogen for complete reduction....and dihydrogen is the LIMITING reagent....

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....

Related questions
See all questions in Limiting Reagent
Impact of this question
1581 views around the world
You can reuse this answer
Creative Commons License