\(n_{CO_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
\(m_{NaOH}=\dfrac{200.9}{100}=18\left(g\right)=>n_{NaOH}=\dfrac{18}{40}=0,45\left(mol\right)\)
PTHH: 2NaOH + CO2 --> Na2CO3 + H2O
Xét tỉ lệ: \(\dfrac{0,45}{2}< \dfrac{0,35}{1}\) => NaOH hết, CO2 dư
PTHH: 2NaOH + CO2 --> Na2CO3 + H2O
_______0,45--->0,225----->0,225____________(mol)
Na2CO3 + CO2 + H2O --> 2NaHCO3
0,125<----0,125--------------->0,25____________(mol)
=> \(\left\{{}\begin{matrix}n_{Na_2CO_3}=0,1\left(mol\right)\\n_{NaHCO_3}=0,25\left(mol\right)\end{matrix}\right.=>\left\{{}\begin{matrix}m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\\m_{NaHCO_3}=0,25.84=21\left(g\right)\end{matrix}\right.\)
