Ta có: \(n_{CO}=\dfrac{1,568}{22,4}=0,07\left(mol\right)\)
\(n_{NaOH}=\dfrac{6,4}{40}=0,16\left(mol\right)\)
PTHH: CO + NaOH ---> HCOONa
Ta thấy: \(\dfrac{0,07}{1}< \dfrac{0,16}{1}\)
=> NaOH dư.
Theo PT: \(n_{HCOONa}=n_{CO}=0,07\left(mol\right)\)
=> \(m_{HCOONa}=0,07.68=4,76\left(g\right)\)