Answer: The mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g
Explanation:
To calculate the number of moles, we use the equation:
.....(1)
Given mass of ammonium sulfate = 20.3 g
Molar mass of ammonium sulfate = 132.14 g/mol
Putting values in equation 1, we get:
The chemical equation for the reaction of ammonia and sulfuric acid follows:
As, sulfuric acid remains unreacted, which means it is an excess reagent and its starting mass cannot be determined from ammonium sulfate.
By Stoichiometry of the reaction:
1 mole of ammonium sulfate is produced by 2 moles of ammonia.
So, 0.154 moles of ammonium sulfate is produced by = of ammonia.
Now, calculating the mass of ammonia from equation 1, we get:
Molar mass of ammonia = 17 g/mol
Moles of ammonia = 0.308 moles
Putting values in equation 1, we get:
Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.
This also means that total mass on the reactant side must be equal to the total mass on the product side.
Let the mass of sulfuric acid be 'x' grams
We are given:
Mass of ammonium sulfate = 20.3 grams
Mass of ammonia = 5.236 grams
Total mass on reactant side = 5.236 + x
Total mass on product side = 20.3 g
So, by applying law of conservation of mass, we get:
Hence, the mass of ammonia is 5.236 g and that of sulfuric acid is 15.064 g
(2) protons (4) positrons
Explanation:
First, calculate the moles of using ideal gas equation as follows.
PV = nRT
or, n =
= (as 1 bar = 1 atm (approx))
= 0.183 mol
As, Density =
Hence, mass of water will be as follows.
Density =
0.998 g/ml =
mass = 3.25 g
Similarly, calculate the moles of water as follows.
No. of moles =
=
= 0.180 mol
Moles of hydrogen = = 0.36 mol
Now, mass of carbon will be as follows.
No. of moles =
0.183 mol =
= 2.19 g
Therefore, mass of oxygen will be as follows.
Mass of O = mass of sample - (mass of C + mass of H)
= 3.50 g - (2.19 g + 0.36 g)
= 0.95 g
Therefore, moles of oxygen will be as follows.
No. of moles =
=
= 0.059 mol
Now, diving number of moles of each element of the compound by smallest no. of moles as follows.
C H O
No. of moles: 0.183 0.36 0.059
On dividing: 3.1 6.1 1
Therefore, empirical formula of the given compound is .
Thus, we can conclude that empirical formula of the given compound is .