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%.
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
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