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

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Answer 1 · 2016-09-04T07:27:06

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

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

What mass of "ammonium sulfate""ammonium sulfate" can be prepared from a 30*g30*g mass of "ammonia""ammonia", and a 196*g196*g mass of "sulfuric acid""sulfuric acid"?