Question

11. Розечин натрия силикату масою 61 г с масовою часткою соли 2% змиша ли с надлишком хлоридною кислотою. Обчисліть масу утвореного осаду.​

Answer 1

Ответ:1,22

Объяснение:61•2%=1,22

Answer 2

Розчин натрію силікату масою 61 г з масовою часткою солі 2% змішали з надлишком хлоридної кислоти. Запитується маса утвореного осаду.

Given Information:

- Mass of sodium silicate solution: 61 g - Mass fraction of salt in the solution: 2%

To calculate the mass of the precipitate formed, we need to determine the reaction that occurs between sodium silicate and hydrochloric acid. Once we know the balanced equation, we can use stoichiometry to find the mass of the precipitate.

Reaction Equation:

The reaction between sodium silicate (Na2SiO3) and hydrochloric acid (HCl) can be represented by the following balanced equation:

Na2SiO3 + 2HCl → 2NaCl + H2SiO3

According to the balanced equation, 1 mole of sodium silicate reacts with 2 moles of hydrochloric acid to produce 2 moles of sodium chloride and 1 mole of silicic acid.

Stoichiometry Calculation:

To calculate the mass of the precipitate formed, we need to determine the number of moles of sodium silicate and hydrochloric acid present in the given solution.

1. Moles of sodium silicate: - Molar mass of sodium silicate (Na2SiO3) = 2(22.99 g/mol)