500.0 g of ammonia react with 51.0 moles of sulfuric acid to produce ammonium sulfate. How many moles of excess sulfuric acid are left over after the reaction is complete?
2 NH3 (aq) + H2SO4(aq) → (NH4)2SO4 (aq)
This is a limiting reagent problem. I know that because BOTH reactants are given. I solve these by solving two simple stoichiometry problems (simple meaning they are NOT limiting reagent problems); one for one reactant and another for the other reactant. Here is an solved example of a simple stoichiometry problem. Just follow the steps. Remember moles = grams/molar mass or moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Post your work if you get stuck and I can help you through it.
To find out how many moles of excess sulfuric acid are left over after the reaction, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a reaction, thus limiting the amount of product that can be formed.
Using the balanced equation, we can determine the stoichiometric ratio between ammonia (NH3) and sulfuric acid (H2SO4). From the equation:
2 NH3 (aq) + H2SO4 (aq) → (NH4)2SO4 (aq)
We can see that for every 2 moles of NH3, 1 mole of H2SO4 is required.
Given:
Mass of ammonia (NH3) = 500.0 g
Molar mass of ammonia (NH3) = 17.03 g/mol
First, let's calculate the number of moles of ammonia (NH3):
moles of NH3 = mass / molar mass
moles of NH3 = 500.0 g / 17.03 g/mol ≈ 29.33 mol
Next, we need to calculate the number of moles of sulfuric acid (H2SO4) required:
moles of H2SO4 = 51.0 mol
Now, let's compare the moles of NH3 to H2SO4 to determine the limiting reactant.
From the balanced equation, it is clear that 2 moles of NH3 react with 1 mole of H2SO4. So, the ratio of NH3 to H2SO4 is 2:1.
The number of moles of H2SO4 to react with the given 29.33 moles of NH3 can be calculated using this ratio:
moles of H2SO4 required = (moles of NH3) / 2
moles of H2SO4 required = 29.33 mol / 2
moles of H2SO4 required = 14.67 mol
We can see that we only need 14.67 moles of H2SO4 to react completely with the given amount of NH3. However, we have 51.0 moles of H2SO4 available.
To determine the amount of excess sulfuric acid, we subtract the moles of H2SO4 required from the moles of H2SO4 available:
excess moles of H2SO4 = moles of H2SO4 available - moles of H2SO4 required
excess moles of H2SO4 = 51.0 mol - 14.67 mol
excess moles of H2SO4 = 36.33 mol
Therefore, there are 36.33 moles of excess sulfuric acid remaining after the reaction is complete.