Question

10 гр хлорида бария взаимодействует с 20 г серной кислотой.найти массу выпавшего остатка

Answer 1

BaCl2 + H2SO4 = BaSO4 + 2 HCl
1. Задача на избыток и недостаток.
2. Находим  n (BaCl2)=10/208= 0,04 моль
3  Находим n (H2SO4) = 20/98 =0,2 моль.
3 Устанавливаем какое из веществ в недостатке.0,04/1=х/0,2, х=0,04*0,2= 0,008 моль- это серная кислота, след., она в избытке (0 0,2 моль)
4. По реакции n( BaSO4) = 0,04 моль
5. Находим массу осадка 0,04*233 = 9,32г
Ответ: 9,32г

Answer 2

Given Information:

- 10 g of barium chloride (BaCl2) reacts with 20 g of sulfuric acid (H2SO4). - We need to find the mass of the precipitate formed.

Balanced Chemical Equation:

To find the mass of the precipitate formed, we first need to write the balanced chemical equation for the reaction between barium chloride and sulfuric acid.

The balanced chemical equation for the reaction is: BaCl2 + H2SO4 → BaSO4 + 2HCl

According to the balanced equation, 1 mole of barium chloride reacts with 1 mole of sulfuric acid to produce 1 mole of barium sulfate and 2 moles of hydrochloric acid.

Molar Mass Calculation:

To calculate the molar mass of barium chloride (BaCl2) and sulfuric acid (H2SO4), we need to sum up the atomic masses of the elements in each compound.

The atomic masses of the elements are as follows: - Barium (Ba): 137.33 g/mol - Chlorine (Cl): 35.45 g/mol - Hydrogen (H): 1.01 g/mol - Sulfur (S): 32.07 g/mol - Oxygen (O): 16.00 g/mol

Using these atomic masses, we can calculate the molar masses of the compounds: - Molar mass of BaCl2 = (1 * 137.33) + (2 * 35.45) = 208.23 g/mol - Molar mass of H2SO4 = (2 * 1.01) + 32.07 + (4 * 16.00) = 98.09 g/mol

Stoichiometry Calculation:

Now, we can use the molar masses and the given masses of the reactants to calculate the limiting reactant and the mass of the precipitate formed.

1. Calculate the number of moles of each reactant: - Moles of BaCl2 = mass / molar mass = 10 g / 208.23 g/mol = 0.048 moles - Moles of H2SO4 = mass / molar mass = 20 g / 98.09 g/mol = 0.204 moles

2. Determine the limiting reactant: The limiting reactant is the one that is completely consumed in the reaction. It is determined by comparing the mole ratios of the reactants in the balanced equation.

From the balanced equation, we can see that the mole ratio of BaCl2 to H2SO4 is 1:1. Therefore, the limiting reactant is BaCl2 because it has fewer moles (0.048 moles) compared to H2SO4 (0.204 moles).

3. Calculate the moles of the precipitate formed: Since the limiting reactant is BaCl2, it will react completely to form the precipitate (BaSO4). According to the balanced equation, 1 mole of BaCl2 produces 1 mole of BaSO4.

Therefore, the moles of BaSO4 formed = moles of BaCl2 = 0.048 moles.

4. Calculate the mass of the precipitate formed: To calculate the mass of the precipitate formed, we need to multiply the moles of BaSO4 by its molar mass.

- Molar mass of BaSO4 = (1 * 137.33) + 32.07 + (4 * 16.00) = 233.39 g/mol

Mass of BaSO4 = moles of BaSO4 * molar mass of BaSO4 = 0.048 moles * 233.39 g/mol = 11.20 g

Answer:

The mass of the precipitate (BaSO4) formed in the reaction between 10 g of barium chloride (BaCl2) and 20 g of sulfuric acid (H2SO4) is 11.20 g.