The chemical equation shows how ammonia reacts with sulfuric acid to produce ammonium sulfate. 2NH3(aq) + H2SO4(aq) (NH4)2SO4(aq) How many grams of ammonium sulfate can be produced if 60.0 mol of sulfuric acid react with an excess of ammonia

7923.6 g of (NH₄)₂SO₄ can be produced by this reaction

Explanation:

The reaction is:

2NH₃ (aq) + H₂SO₄(aq)  → (NH₄)₂SO₄(aq)

In this reaction ratio is 1:1.

As the ammonia is in excess, the limiting reagent is the sulfuric acid.

So 1 mol of sulfuric can produce 1 mol of sulfate

Then, 60 moles of sulfuric must produce 60 moles of sulfate.

We convert the moles to mass:

60 mol . 132.06 g / 1mol = 7923.6 g

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation: