Question

СРОЧНО!!!!! Рассчитайте количество вещества (моль) и массу Сульфита бария, образующегося при проведении следующих превращений: H2SO4→ BaSO4 → , если было взято 164 г сернистой кислоты H2SO4.

Answer 1

Так как сернистая кислота - это H2SO3, и именно ее соли называются сульфитами - исправьте ошибку в формуле кислоты.
H2SO3 + Ba(OH)2 = BaSO3↓ + 2H2O
∨(H2SO3) = 164 (г)/82 (г/моль) = 2 моль
По уравнению реакции видно, что количества сернистой кислоты и сульфита бария равны, значит ∨(BaSO3) = 2 моль.
Масса BaSO3 = 2 (моль) х 217 (г/моль) = 434 г

Answer 2

Calculation of the Number of Moles and Mass of Barium Sulfite (BaSO4)

To calculate the number of moles and mass of barium sulfite (BaSO4) formed from the given transformation, we need to use the molar mass of sulfuric acid (H2SO4) and the stoichiometry of the reaction.

The balanced equation for the reaction is as follows: H2SO4 → BaSO4

From the equation, we can see that one mole of sulfuric acid (H2SO4) reacts to form one mole of barium sulfite (BaSO4).

To calculate the number of moles of barium sulfite (BaSO4) formed, we need to determine the number of moles of sulfuric acid (H2SO4) used. We can use the molar mass of sulfuric acid to convert the given mass of sulfuric acid (H2SO4) into moles.

The molar mass of sulfuric acid (H2SO4) is calculated as follows: 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol

Given that the mass of sulfuric acid (H2SO4) is 164 g, we can calculate the number of moles using the following equation: Number of moles = Mass / Molar mass

Number of moles of sulfuric acid (H2SO4) = 164 g / 98.09 g/mol = 1.67 mol

Since the stoichiometry of the reaction is 1:1, the number of moles of barium sulfite (BaSO4) formed is also 1.67 mol.

To calculate the mass of barium sulfite (BaSO4) formed, we can use the molar mass of barium sulfite (BaSO4), which is calculated as follows: 137.33 g/mol + 32.07 g/mol + 4(16.00 g/mol) = 233.39 g/mol

Mass of barium sulfite (BaSO4) = Number of moles × Molar mass Mass of barium sulfite (BaSO4) = 1.67 mol × 233.39 g/mol = 389.53 g

Therefore, the number of moles of barium sulfite (BaSO4) formed is 1.67 mol, and the mass of barium sulfite (BaSO4) formed is 389.53 g.

Please note that the above calculations assume the reaction goes to completion and there are no other limiting factors or side reactions involved.