Đáp án: $12,6g$
$n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)$
$n_{OH}=n_{NaOH}=C_M.V=1.0,15=0,15(mol)$
$Ta$ $có:$ $T=\dfrac{n_{OH^-}}{n_{CO_2}}=$ $\dfrac{0,15}{0,15}=1$ $→CO_2$ dư
$Pt:$ $NaOH+CO_2→NaHCO_3$
$n_{NaHCO_3}=n_{NaOH}=n_{CO_2}=0,15(mol)$
$⇒m_{NaHCO_3}=n.M=0,15.84=12,6(g)$
BẠN THAM KHẢO NHA!!!