Question

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)`

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....