What amount of ammonium chloride can be obtained by reacting 75 grams of ammonia containing

| September 1, 2021 | education

What amount of ammonium chloride can be obtained by reacting 75 grams of ammonia containing 10% impurities with hydrochloric acid?

1. Let’s determine how much pure ammonia, without impurities:
Mass fraction of ammonia w (NH3) = 100% -10% = 90%
w (in-va) = (m (in-va) / m (solution)) * 100%.
Let us express the mass of a substance from this expression:
m (in-va) = (w (in-va) * m (solution)) / 100%.
m (NH3) = (w (NH3) * m (solution)) / 100%.
Let’s calculate:
m (NH3) = (w (NH3) * m (solution)) / 100% = 90 * 75/100 = 67.5 g.
2. Let us write down the reaction equation:
NH3 + HCl = NH4Cl.
3. According to the reaction equation, the amount of substance:
ν (NH3) = ν (NH4Cl).
Knowing that ν (in-va) = m (in-va) / M (in-va), we get:
m (NH3) / M (NH3) = m (NH4Cl) / M (NH4Cl).
Let’s define the molar masses:
M (NH3) = 14 + 1 * 4 = 17 g / mol.
M (NH4Cl) = 14 + 1 * 4 + 35.5 = 53.5 g / mol.
Let us express the mass of the ammonium salt:
m (NH4Cl) = m (NH3) * M (NH4Cl) / M (NH3).
Substituting the numerical values, we get:
m (NH4Cl) = m (NH3) * M (NH4Cl) / M (NH3) = 67.5 * 53.5 / 17 = 212.4 g.
Answer: You can get 212.4 g of ammonium chloride.



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