Question

Помогите решить на зачет!!! заранее спасибо))) при взаимодействии 16гр. раствора серной кислоты с избытком раствора хлорида бария ,образовалось 5,7гр. осадка. рассчитать массовую долю серной кислоты в исходном растворе.

Answer 1

BaCl2 + H2SO4 = BaSO4 + 2HCl
M(BaSO4) = 233 г/моль
n(BaSO4) = 5.7 /233 =0.024 моль
M(H2SO4) = 98 г/моль
m(H2SO4) = 0.024 *98 =2.352 г
w(H2SO4) = 2.352/16 * 100% = 14.7 %

Answer 2

Solution to the Problem

To solve this problem, we can use the concept of chemical reactions and stoichiometry. The reaction between sulfuric acid (H2SO4) and barium chloride (BaCl2) can be represented as follows:

H2SO4 + BaCl2 → BaSO4 + 2HCl

From the balanced chemical equation, we can see that 1 mole of sulfuric acid reacts with 1 mole of barium chloride to produce 1 mole of barium sulfate and 2 moles of hydrochloric acid.

Calculation Steps

1. Calculate the molar mass of BaSO4 (barium sulfate) and H2SO4 (sulfuric acid). 2. Use the molar mass of BaSO4 to convert the mass of the precipitate (5.7g) to moles. 3. Use the stoichiometry of the reaction to find the moles of H2SO4 that reacted. 4. Calculate the mass of H2SO4 in the original solution. 5. Finally, calculate the mass percent of H2SO4 in the original solution.

Molar Mass Calculation

The molar mass of BaSO4: - Ba: 137.33 g/mol - S: 32.07 g/mol - O (4 atoms): 16.00 g/mol * 4 = 64.00 g/mol - Total molar mass of BaSO4 = 137.33 + 32.07 + 64.00 = 233.40 g/mol

The molar mass of H2SO4: - H (2 atoms): 1.01 g/mol * 2 = 2.02 g/mol - S: 32.07 g/mol - O (4 atoms): 16.00 g/mol * 4 = 64.00 g/mol - Total molar mass of H2SO4 = 2.02 + 32.07 + 64.00 = 98.09 g/mol

Moles of BaSO4 Formed

Using the molar mass of BaSO4: - Moles of BaSO4 = Mass of BaSO4 / Molar mass of BaSO4 - Moles of BaSO4 = 5.7g / 233.40 g/mol = 0.0244 moles

Moles of H2SO4 Reacted

From the balanced chemical equation, 1 mole of BaSO4 is formed from 1 mole of H2SO4: - Moles of H2SO4 = Moles of BaSO4

Mass of H2SO4 in the Original Solution

Using the molar mass of H2SO4: - Mass of H2SO4 = Moles of H2SO4 * Molar mass of H2SO4 - Mass of H2SO4 = 0.0244 moles * 98.09 g/mol = 2.39g

Mass Percent of H2SO4 in the Original Solution

The mass percent can be calculated using the formula: Mass percent = (Mass of H2SO4 / Total mass of solution) * 100%

Given that the mass of the precipitate is 5.7g, we can assume the total mass of the solution is close to the mass of the precipitate.

- Mass percent = (2.39g / 5.7g) * 100% = 41.93%

Therefore, the mass percent of sulfuric acid in the original solution is approximately 41.93%.