What volume of carbon dioxide will be released when 0.25 mol of sodium carbonate interacts

What volume of carbon dioxide will be released when 0.25 mol of sodium carbonate interacts with an excess of hydrochloric acid?

Let’s implement the solution:
1. We write down the equation according to the problem statement:
Y = 0.25 mol. X l. -?
Na2CO3 + 2HCl = 2NaCl + CO2 + H2O – carbon dioxide is released during ion exchange;
2. Calculations by formulas:
M (Na2CO3) = 105.8 g / mol.
3. Let’s calculate the amount of CO2:
Y (Na2CO3) = 0.25 mol;
Y (CO2) = 0.25 mol since the amount of these substances according to the equation is 1 mol.
4. Determine the volume of the product:
V (CO2) = 0.25 * 22.4 = 5.6 liters.
Answer: during the reaction, 5.6 liters of carbon dioxide was formed.