What mass of #"ammonium sulfate"# can be prepared from a #30*g# mass of #"ammonia"#, and a #196*g# mass of #"sulfuric acid"#?

1 Answer
Sep 4, 2016

Answer:

#2NH_3(aq) + H_2SO_4(aq) rarr (NH_4)_2SO_4(aq)#

Explanation:

We have a balanced chemical equation that represents the neutralization of ammonia by sulfuric acid.

#"Moles of ammonia"# #=# #(30*g)/(17*g*mol^-1)# #=# #1.76*mol#.

#"Moles of vitriol"# #=# #(196*g)/(98.08*g*mol^-1)# #=# #2.00*mol#.

Now clearly, given the 2:1 stoichiometry, ammonia is the limiting reagent, and sulfuric is in excess. (What do I mean by 2:1 stoichiometry?)

Given #1.76# #mol# ammonia, we can make #(1.76*mol)/2# ammonium sulfate;

i.e. #(1.76*cancel(mol))/2xx132.1*g*cancel(mol^-1)# #=# #??g#