Question

К раствору массой 200 г с массовой долей силиката натрия 6,1% прилили соляную кислоту. Вычислите массу образовавшегося осадка.

Answer 1

Na₂SiO₃+ 2HCI->2NaCI+ H₂SiO₃↓
1) Найдем  m ( Na₂SiO₃)=  200* 0,061= 12,2 г
V(ню)= m:M= 12, 2  :122 =0, 1 моль
M(Na₂SiO₃)= 23*2+ 28+16*3= 122 г. моль
2) По уравнению: V(H₂SiO₃)=V(Na₂SiO₃)= 0,1 моль
3) Найдем массу осадка:
m(H₂SiO₃)= V* M= 0,1*78=7,8 г
M ( H₂SiO₃)=1*2+28+16*3= 78 г. моль 
Ответ: 7, 8 г. 

Answer 2

Calculating the mass of the precipitate formed

The question provides the following information: - Mass of the solution = 200 g - Mass fraction of sodium silicate (Na2SiO3) in the solution = 6.1%

To calculate the mass fraction of sodium silicate: - Mass fraction = Mass of solute / Total mass - Mass of sodium silicate = 6.1% of 200 g = 0.061 × 200 = 12.2 g (the mass of sodium silicate in the solution)

When hydrochloric acid (HCl) is added to the solution, sodium silicate reacts with HCl to form silicon dioxide (SiO2) precipitate. The balanced chemical equation for the reaction is:

`Na2SiO3 + 2HCl → SiO2 + 2NaCl`

Using the balanced equation: - Moles of sodium silicate = Mass of sodium silicate / Molar mass of sodium silicate = 12.2 g / 122 g/mol = 0.1 mol - Moles of silicon dioxide formed = Moles of sodium silicate = 0.1 mol - Mass of silicon dioxide formed = Moles of silicon dioxide × Molar mass of silicon dioxide = 0.1 mol × 60 g/mol = 6 g

Therefore, the mass of the silicon dioxide precipitate formed is 6 g.