Answer:
72.66%
Explanation:
NH₄Cl reacts in presence of NaOH producing ammonia, NH₃, as follows:
NH₄Cl + NaOH → NaCl + NH₃ + H₂O
The residual NaOH reacts with H₂SO₄ as follows:
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O.
The equivalent-gram of H₂SO₄ are:
16mL * 0.1N * 1.06 = 1.696mEq.
As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:
1.696mEq * (100mL / 10mL) = 16.96mEq.
The mEq of NaOH you add in the first are:
25mL * 0.95mEq = 23.75mEq
That means the NaOH that reacts = moles of NH₄Cl is:
23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =
6.79x10⁻³ moles NH₄Cl
In grams (Using molar mass NH₄Cl = 53.5g/mol):
6.79x10⁻³ moles NH₄Cl * (53.5g / mol) =
0.3633g of NH₄Cl are in the original mixture.
% is:
0.3633g/ 0.5g * 100 = 72.66%
The percentage of pure NH₄Cl in the mixture = 72.66%
Given data :
0.5 gm of mixture of NH₄Cl and NaCI
25 ml of 0.95 N NaOH
Determine the % of pure NH₄Cl in the mixture
Given that the reaction equation of the mixture is ;
NH₄Cl + NaOH ----> NaCl + NH₃ + H₂O
And the residual solution reaction with H₂SO₄ is
2NaOH + H₂SO₄ ----> Na₂SO₄ + 2H₂O.
∴ H₂SO₄ ( equivalent gram ) = 16mL * 0.1N * 1.06 = 1.696
but the equivalent gram of H₂SO₄ required to neutralize NaOH = 1.696 * ( 100 mL / 10 mL ) = 16.96
NaOH⁻ = 25mL * 0.95 = 23.75mEq ( from the first reaction equation )
next step : Determine the number of moles of NH₄Cl
moles of NH₄Cl = 23.75- 16.96 = 6.79 * 10⁻³
Final step : Calculate the % of pure NH₄CI in the mixture
moles of NH₄Cl = 6.79 * 10⁻³
molar mass of NH₄Cl = 53.5g/mol
∴ mass of NH₄Cl = 6.79 * 10⁻³ * 53.5 = 0.3633g
% of pure NH₄CI in the mixture = ( 0.3633 / 0.5 ) * 100 = 72.66%
Hence we can conclude that The percentage of pure NH₄Cl in the mixture is 72.66%
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