$n_{CO_2}=\dfrac{1,12}{22,4}=0,05\ (mol)$
$n_{NaOH}=\dfrac{160.1}{40.100}=0,04\ (mol)$
$\to T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,04}{0,05}=0,8<1$
$\to$ Phản ứng tạo muối $NaHCO_3$, $NaOH$ hết
PTHH: $CO_2+NaOH\to NaHCO_3$
Theo PT: $n_{NaHCO_3}=n_{NaOH}=0,04\ (mol)$
$\to m_{NaHCO_3}=0,04.84=3,36\ (g)$