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K₂CO₃+2HCl →2KCl + CO₂ + H₂O

ta có: 7,437l CO₂

1. n_CO2= 7,437/22,4≈ 0,3mol

n​K2CO3= n_CO2= 0,3mol

m_K2CO3= 0,3 . (39.2+16.3+12)= 41,4g 

⇒ m= 41,4g

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