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

Question

Peyton Adams · Accepted Answer

#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#

Avery Adams · Answer

undefined To find the mass of ammonium sulfate that can be prepared, you first need to determine the molar masses of ammonia (NH₃) and sulfuric acid (H₂SO₄), then use stoichiometry to calculate the mass of ammonium sulfate (NH₄)₂SO₄ formed.

1. Calculate the molar masses:
   - Molar mass of NH₃ (ammonia) = 14.01 g/mol (N) + 3(1.008 g/mol) (H) = 17.03 g/mol
   - Molar mass of H₂SO₄ (sulfuric acid) = 2(1.008 g/mol) (H) + 32.07 g/mol (S) + 4(16.00 g/mol) (O) = 98.09 g/mol

2. Determine the stoichiometry of the reaction:
   The balanced chemical equation for the reaction between ammonia and sulfuric acid to form ammonium sulfate is:
   2NH₃ + H₂SO₄ → (NH₄)₂SO₄

3. Calculate the amount of each reactant used:
   - moles of NH₃ = mass / molar mass = 30 g / 17.03 g/mol ≈ 1.76 mol
   - moles of H₂SO₄ = mass