Determine the amount of gas substance obtained by the interaction of 200 g of 6.9% sodium carbonate solution

Determine the amount of gas substance obtained by the interaction of 200 g of 6.9% sodium carbonate solution with an excess of hydrochloric acid according to the reaction equation: Na1CO3 + 2HCL = 2NaCl + CO2 + H2O

Given:
m solution (Na2CO3) = 200g
w (Na2CO3) = 6.9% = 0.069
Find: n (CO2)
Decision:
Na2CO3 + 2HCL = 2NaCl + CO2 + H2O
m in-va (Na2CO3) = w * m solution = 0.069 * 200g = 13.8 g
n (Na2CO3) = m / M = 13.8 g / 142 g / mol = 0.097 mol
n (Na2CO3): n (CO2) = 1: 1
n (CO2) = 0.097 mol
Answer: 0.097 mol