0.500g of impure ammonium chloride is warmed with an excess of NaOH solution.the AMMONIA LIBERATED IS ABSORBED 25.0 CM OF 0.200 ML/L OF SULPHURIC ACID ,THE excess of sulphuric acid require 5.64cm ,0.200mol/l of NaOH solution of titration. calculate the percentage of ammonia chloride in the original sample
First, we need to determine the amount of sulphuric acid that reacted with ammonia. We know that after the reaction, 5.64 cm³ of 0.2 mol/L NaOH solution was required to neutralize the excess sulphuric acid. We can calculate the amount of acid that reacted with NaOH using the formula:
moles of acid = volume x concentration
moles of H₂SO₄ that reacted with NaOH = 5.64 cm³ x 0.2 mol/L = 1.128 mmol of H₂SO₄
Since the initial volume of sulphuric acid was 25.0 cm³ and the concentration was 0.2 mol/L, the initial amount of sulphuric acid was:
Initial moles of H₂SO₄ = 25.0 cm³ x 0.2 mol/L = 5.0 mmol
So to find the moles of H₂SO₄ that reacted with ammonia, we subtract the moles that reacted with NaOH:
moles of H₂SO₄ that reacted with NH₃ = 5.0 mmol - 1.128 mmol = 3.872 mmol
Now we need to determine the moles of ammonium chloride that reacted to form this ammonia. Since ammonium chloride reacts in a 1:1 ratio with sulphuric acid to form ammonia:
NH₄Cl + H₂SO₄ → NH₃ + HCl + H₂SO₄
moles of NH₄Cl = moles of H₂SO₄ that reacted with NH₃ = 3.872 mmol
To find the mass of ammonium chloride, we use the formula:
mass = moles x molar mass
The molar mass of NH₄Cl is 53.49 g/mol. Therefore,
mass of NH₄Cl = 3.872 mmol x 53.49 g/mol = 0.207 g
Finally, we can calculate the percentage of ammonium chloride in the original sample:
percentage of NH₄Cl = (mass of NH₄Cl / total mass) x 100%
percentage of NH₄Cl = (0.207 g / 0.500 g) x 100% = 41.4%
So the percentage of ammonium chloride in the original sample is 41.4%.