300.0 g of ammonia react with 36.0 moles of sulfuric acid to produce ammonium sulfate. How many moles of excess sulfuric acid are left over after the reaction is complete?
sulfuric acid + ammonia -----> ammonium sulfate
Therefore,
H2SO4 + 2NH3 --> (NH4)2SO4
http://www.ausetute.com.au/sulfacid.html
This website will help you do the rest of the problem:
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/limiting.htm
To determine how many moles of excess sulfuric acid are left over, we need to compare the amount of ammonia and sulfuric acid used in the reaction.
First, we need to convert the mass of ammonia to moles. To do this, we will use the molar mass of ammonia (NH3), which is approximately 17.03 g/mol.
moles of ammonia = mass of ammonia / molar mass of ammonia
moles of ammonia = 300.0 g / 17.03 g/mol
moles of ammonia ≈ 17.62 mol
Next, we compare the moles of ammonia to the moles of sulfuric acid to determine the limiting reagent. The balanced chemical equation for the reaction is:
2NH3 + H2SO4 -> (NH4)2SO4
From the equation, we can see that the stoichiometric ratio between ammonia and sulfuric acid is 2:1. This means that for every 2 moles of ammonia, we need 1 mole of sulfuric acid.
moles of sulfuric acid needed = moles of ammonia / 2
moles of sulfuric acid needed ≈ 17.62 mol / 2
moles of sulfuric acid needed ≈ 8.81 mol
Since we have an excess of sulfuric acid (36.0 moles), we subtract the moles of sulfuric acid needed from the total moles of sulfuric acid to find the moles of excess sulfuric acid.
moles of excess sulfuric acid = total moles of sulfuric acid - moles of sulfuric acid needed
moles of excess sulfuric acid = 36.0 mol - 8.81 mol
moles of excess sulfuric acid ≈ 27.19 mol
Therefore, approximately 27.19 moles of sulfuric acid are left over after the reaction is complete.