Question

Имеется раствор, содержащий одновременно серную и азотную кислоты. Для полной нейтрализации 10,0 г этого раствора расходуется 12,5 см3 раствора КОН (w=19процентов, р=1,18 г/см3). При добавлении к 20,0 г этой же смеси раствора кислот избытка хлорида бария выпадает 4,66 г осадка. Найдите массовые доли кислот в смеси.

Answer 1

Пишем реакции
#1 H2SO4 + BaCl2 = BaSO4 + 2HCl
#2 H2SO4 + 2KOH = K2SO4 + H2O
#3 HNO3 + KOH = KNO3 + H2O
Считаем количество сульфата бария m/M = 4.66/233 = 0.02 моль.
Количество серной кислоты по #1 такое же = 0,02 моль. Значит масса = 0,02*98 = 1,96 грамм.
Если в растворе 20 г, серной кислоты 1,96 грамма, то в растворе 10 г, соответственно 1,96/2 = 0,98 г (0,01 моль)
По #2 количество КОН в 2 раза больше чем серной кислоты = 0,02 моль. Масса n*M = 0.02*56 = 1.12 грамма.
Считаем сколько у нас есть всего КОН. Масса раствора = 12,5*1,18 = 14,75 грамм. Масса КОН = 14,75*0,19 = 2,8 г.
КОН который пойдет на реакцию с азотной кислотой = 2,8-1.12 = 1.68 г. Количество его 1.68/56 = 0,03 моль.
По реакции #3 такое же количество и азотной кислоты, значит масса ее n*M = 0.03*63 = 1.89 г.
Ну и наконец массовые доли
H2SO4 = 0.98/10 = 0,098 или 9.8%
HNO3 = 1.89/10 = 0.189 или 18.9%

Answer 2

Calculation of Acid Mass Fractions in the Mixture

To find the mass fractions of sulfuric acid (H2SO4) and nitric acid (HNO3) in the mixture, we can use the information provided.

Given: - Mass of the solution = 10.0 g - Volume of KOH solution required for complete neutralization = 12.5 cm^3 - Concentration of KOH solution (w) = 19% - Density of KOH solution (p) = 1.18 g/cm^3 - Mass of the acid mixture added = 20.0 g - Mass of precipitate formed (BaCl2) = 4.66 g

To calculate the mass fractions, we need to determine the moles of sulfuric acid and nitric acid in the mixture.

1. Moles of KOH used for neutralization: The molar mass of KOH (potassium hydroxide) is 56.11 g/mol. The concentration of KOH solution (w) is given as 19%, which means there are 19 g of KOH in 100 g of the solution. Therefore, the mass of KOH in the solution is (19/100) * 12.5 g. The moles of KOH used can be calculated using the formula: moles = mass / molar mass.

2. Moles of H2SO4 and HNO3: Since the volume of KOH solution used for neutralization is not given, we cannot directly calculate the moles of H2SO4 and HNO3. However, we can use the stoichiometry of the reaction to determine the moles.

The balanced chemical equation for the neutralization reaction between KOH and H2SO4 is: 2 KOH + H2SO4 -> K2SO4 + 2 H2O

From the equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. Therefore, the moles of H2SO4 can be calculated as: moles of H2SO4 = (2 * moles of KOH used)

Similarly, the balanced chemical equation for the reaction between KOH and HNO3 is: KOH + HNO3 -> KNO3 + H2O

From the equation, we can see that 1 mole of KOH reacts with 1 mole of HNO3. Therefore, the moles of HNO3 can be calculated as: moles of HNO3 = moles of KOH used

3. Moles of BaCl2: The molar mass of BaCl2 (barium chloride) is 208.23 g/mol. The moles of BaCl2 can be calculated using the formula: moles = mass / molar mass.

4. Moles of H2SO4 and HNO3 in the mixture: Since the moles of KOH used for neutralization are known, we can calculate the moles of H2SO4 and HNO3 using the stoichiometry of the reaction.

From the balanced chemical equation between KOH and H2SO4, we know that 2 moles of KOH react with 1 mole of H2SO4. Therefore, the moles of H2SO4 in the mixture can be calculated as: moles of H2SO4 = (moles of KOH used) / 2

From the balanced chemical equation between KOH and HNO3, we know that 1 mole of KOH reacts with 1 mole of HNO3. Therefore, the moles of HNO3 in the mixture can be calculated as: moles of HNO3 = moles of KOH used

5. Mass fractions of H2SO4 and HNO3: The mass fraction of a component in a mixture is calculated by dividing the mass of the component by the total mass of the mixture.

The mass fraction of H2SO4 can be calculated as: mass fraction of H2SO4 = (moles of H2SO4 * molar mass of H2SO4) / mass of the acid mixture added

The mass fraction of HNO3 can be calculated as: mass fraction of HNO3 = (moles of HNO3 * molar mass of HNO3) / mass of the acid mixture added

Now, let's perform the calculations using the given values.

Calculation Steps:

1. Calculate the moles of KOH used for neutralization: - Mass of KOH in the solution = (19/100) * 12.5 g - Moles of KOH used = (mass of KOH in the solution) / molar mass of KOH

2. Calculate the moles of H2SO4 and HNO3: - Moles of H2SO4 = 2 * (moles of KOH used) - Moles of HNO3 = moles of KOH used

3. Calculate the moles of BaCl2: - Moles of BaCl2 = mass of precipitate formed / molar mass of BaCl2

4. Calculate the moles of H2SO4 and HNO3 in the mixture: - Moles of H2SO4 in the mixture = (moles of KOH used) / 2 - Moles of HNO3 in the mixture = moles of KOH used

5. Calculate the mass fractions of H2SO4 and HNO3: - Mass fraction of H2SO4 = (moles of H2SO4 * molar mass of H2SO4) / mass of the acid mixture added - Mass fraction of HNO3 = (moles of HNO3 * molar mass of HNO3) / mass of the acid mixture added

Let's perform the calculations:

1. Calculate the moles of KOH used for neutralization: - Mass of KOH in the solution = (19/100) * 12.5 g = 2.375 g - Molar mass of KOH = 56.11 g/mol - Moles of KOH used = (2.375 g) / (56.11 g/mol) = 0.0423 mol

2. Calculate the moles of H2SO4 and HNO3: - Moles of H2SO4 = 2 * (0.0423 mol) = 0.0846 mol - Moles of HNO3 = 0.0423 mol

3. Calculate the moles of BaCl2: - Mass of BaCl2 = 4.66 g - Molar mass of BaCl2 = 208.23 g/mol - Moles of BaCl2 = (4.66 g) / (208.23 g/mol) = 0.0224 mol

4. Calculate the moles of H2SO4 and HNO3 in the mixture: - Moles of H2SO4 in the mixture = 0.0846 mol - Moles of HNO3 in the mixture = 0.0423 mol

5. Calculate the mass fractions of H2SO4 and HNO3: - Mass of the acid mixture added = 20.0 g - Molar mass of H2SO4 = 98.09 g/mol - Molar mass of HNO3 = 63.01 g/mol - Mass fraction of H2SO4 = (0.0846 mol * 98.09 g/mol) / 20.0 g = 0.4169 - Mass fraction of HNO3 = (0.0423 mol * 63.01 g/mol) / 20.0 g = 0.1340

Therefore, the mass fractions of sulfuric acid (H2SO4) and nitric acid (HNO3) in the mixture are approximately 41.69% and 13.40%, respectively.

Please note that these calculations are based on the given information and assumptions made.