a,
$n_{CO_2}=\dfrac{3,163}{22,4}=0,141(mol)$
$n_{NaOH}=\dfrac{12,8}{40}=0,32(mol)$
$2NaOH+CO_2\to Na_2CO_3+H_2O$
Ta có $\dfrac{0,32}{2}>\dfrac{0,141}{1}$
Nên $NaOH$ dư
$n_{NaOH\text{pứ}}=2n_{CO_2}=0,141.2=0,282(mol)$
$\Rightarrow n_{NaOH\text{dư}}=0,32-0,282=0,038(mol)$
$\Rightarrow m_{NaOH\text{dư}}=0,038.40=1,52g$
b,
$n_{Na_2CO_3}=n_{CO_2}=0,141(mol)$
$\Rightarrow m_{Na_2CO_3}=0,141.106=14,946g$