How many grams of #H_2# must react, if the reaction needs to produced 63.5g of #NH_3#?

1 Answer
Feb 6, 2017

Answer:

We need approx. #12*g# of #"dihydrogen gas........."#

Explanation:

As a preliminary we require a stoichiometrically balanced equation:

#1/2N_2(g) + 3/2H_2(g) rightleftharpoonsNH_3(g)#

Is this balanced?

#"Moles of ammonia"=(63.5*g)/(17.01*g*mol^-1)=3.73*mol.#

Given the stoichiometric equation, clearly, we need #3/2# #"equiv"# of #"dihydrogen gas:"#

#"i.e."# #3/2xx3.73*molxx2.02*g*mol^-1=11.3*g#

How much dinitrogen do we need?