*Phương pháp nối tiếp
Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,75\cdot0,2=0,15\left(mol\right)\\n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo 2 muối
PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
0,15___0,075______0,075 (mol)
\(Na_2CO_3+H_2O+CO_2\rightarrow2NaHCO_3\)
0,025__________0,025_______0,05 (mol)
Theo các PTHH: \(\left\{{}\begin{matrix}n_{Na_2CO_3}=0,075-0,025=0,05\left(mol\right)\\n_{NaHCO_3}=0,05\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{muối}=0,05\cdot106+0,05\cdot84=9,5\left(g\right)\)
`n_(CO_2) = (2,24)/(22,4)=0,1(mol)`
`n_(NaOH)=0,75 . 0,2=0,15`
`=> 1 < (n_(CO_2))/(n_(NaOH)) <2`
`=>` Tạo 2 muối: `NaHCO_3` và `Na_2CO_3`.
`CO_2+NaOH->NaHCO_3`
....`x`........`x`..........`x`
`CO_2+2NaOH->Na_2CO_3+H_2O`
....`y`.........`2y`.........`y`..............`y`
`=> {(x+y=0.1),(x+2y=0.15):} <=> x=y=0,05`
`=> m_(\text{muối})=m_(NaHCO_3)+m_(Na_2CO_3)`
`=0,05.84+0,05.106=9,5(g)`