Question

Решите задачку пожалуйста : Раствор карбоната натрия обработали 25г 15% раствора хлороводродной кислоты. Какой объем газа выделился после реакции?

Answer 1

В приложении........................

Answer 2

Task Solution:

To solve this problem, we can use the stoichiometry of the reaction between sodium carbonate (Na2CO3) and hydrochloric acid (HCl). The balanced chemical equation for the reaction is:

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

From the equation, we can see that 1 mole of sodium carbonate reacts with 2 moles of hydrochloric acid to produce 1 mole of carbon dioxide gas.

Given: - Mass of sodium carbonate (Na2CO3) = 25g - Concentration of hydrochloric acid (HCl) = 15%

Calculations:

1. Calculate the number of moles of sodium carbonate (Na2CO3): - The molar mass of Na2CO3 = 22.99*2 + 12.01 + 16*3 = 105.99 g/mol - Number of moles = Mass / Molar mass = 25g / 105.99 g/mol

2. Calculate the number of moles of hydrochloric acid (HCl): - Concentration of HCl = 15% - This means there are 15g of HCl in 100g of solution. - Number of moles of HCl = Mass / Molar mass = 15g / 36.46 g/mol

3. Determine the limiting reagent: - Compare the moles of Na2CO3 and HCl to find the limiting reagent.

4. Calculate the moles of carbon dioxide (CO2) produced: - Since 1 mole of Na2CO3 produces 1 mole of CO2, the moles of CO2 produced will be equal to the moles of Na2CO3 or the moles of HCl, depending on the limiting reagent.

5. Convert moles of CO2 to volume at STP (Standard Temperature and Pressure): - 1 mole of any gas at STP occupies 22.4 liters.

Answer:

The volume of gas produced can be calculated using the steps above. Let's proceed with the calculations and find the volume of gas produced.