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%
