Question

Сколько грамм осадка образуется при взаимодействии 2.07 грамм хлорида бария с 5 молями сульфата натрия. Помогите пожалуйста, подробно только

Answer 1

Как-то так)
если что-то непонятно, задавай вопросы))

Answer 2

Calculation of Precipitate Formation

To calculate the amount of precipitate formed when 2.07 grams of barium chloride reacts with 5 moles of sodium sulfate, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed.

First, let's calculate the molar mass of barium chloride (BaCl2) and sodium sulfate (Na2SO4): - Molar mass of BaCl2 = atomic mass of Ba (137.33 g/mol) + 2 * atomic mass of Cl (35.45 g/mol) = 137.33 g/mol + 2 * 35.45 g/mol = 208.23 g/mol - Molar mass of Na2SO4 = 2 * atomic mass of Na (22.99 g/mol) + atomic mass of S (32.07 g/mol) + 4 * atomic mass of O (16.00 g/mol) = 2 * 22.99 g/mol + 32.07 g/mol + 4 * 16.00 g/mol = 142.04 g/mol

Next, let's calculate the number of moles of barium chloride and sodium sulfate: - Moles of BaCl2 = mass of BaCl2 / molar mass of BaCl2 = 2.07 g / 208.23 g/mol = 0.00994 mol - Moles of Na2SO4 = 5 mol

The balanced chemical equation for the reaction between barium chloride and sodium sulfate is: BaCl2 + Na2SO4 -> BaSO4 + 2NaCl

From the balanced equation, we can see that 1 mole of barium chloride reacts with 1 mole of sodium sulfate to form 1 mole of barium sulfate. Therefore, the stoichiometric ratio between barium chloride and barium sulfate is 1:1.

Since the stoichiometric ratio is 1:1, the moles of barium sulfate formed will be equal to the moles of barium chloride used. Therefore, the moles of barium sulfate formed will be 0.00994 mol.

Finally, let's calculate the mass of barium sulfate formed: - Mass of BaSO4 = moles of BaSO4 * molar mass of BaSO4 = 0.00994 mol * (atomic mass of Ba (137.33 g/mol) + atomic mass of S (32.07 g/mol) + 4 * atomic mass of O (16.00 g/mol)) = 0.00994 mol * (137.33 g/mol + 32.07 g/mol + 4 * 16.00 g/mol) = 0.00994 mol * 233.38 g/mol = 2.32 g

Therefore, when 2.07 grams of barium chloride reacts with 5 moles of sodium sulfate, approximately 2.32 grams of barium sulfate will be formed.

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