`a)`
\(Na_2CO_3+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+2NaOH\) (1)
\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\) (2)
`b)`\(n_{Na_2CO_3}=\dfrac{m}{106}\left(mol\right)\);\(n_{Ca\left(OH\right)_2}=\dfrac{m}{74}\left(mol\right)\);\(n_{NaCl}=\dfrac{m}{58,5}\left(mol\right)\)
\(\left(1\right)\Rightarrow m_{NaOH}=40.2.m_{Na_2CO_3}=\dfrac{80m}{106}=\dfrac{40m}{53}\left(g\right)\)
\(\left(2\right)\Rightarrow m_{NaOH}=40.n_{NaCl}=\dfrac{40m}{58,5}\left(g\right)\)
\(\Rightarrow\) Phản ứng (1) Có thể điều chế NaOH nhiều hơn \(\left(\dfrac{40m}{53}>\dfrac{40m}{58,5}\right)\)
a) `Na_2CO_3 + Ca(OH)_2 -> CaCO_3 + 2NaOH`
$2NaCl + 2H_2O \xrightarrow[cmn]{đpdd} 2NaOH + Cl_2 + H_2$
b) Đặt \(m_{Ca\left(OH\right)_2}=m_{Na_2CO_3}=m_{NaCl}=a\left(g\right)\)
`=>` \(\left\{{}\begin{matrix}n_{Ca\left(OH\right)_2}=\dfrac{a}{74}\left(mol\right)\\n_{Na_2CO_3}=\dfrac{a}{106}\left(mol\right)\\n_{NaCl}=\dfrac{a}{58,5}\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca\left(OH\right)_2+Na_2CO_3\rightarrow CaCO_3\downarrow+2NaOH\)
Xét tỉ lệ: \(\dfrac{a}{74}>\dfrac{a}{106}\Rightarrow Ca\left(OH\right)_2\) dư, `Na_2CO_3` hết
Theo PT: \(n_{NaOH}=2n_{Na_2CO_3}=\dfrac{a}{53}\left(mol\right)\)
PTHH: \(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+Cl_2+H_2\)
Theo PT: \(n_{NaOH}=n_{NaCl}=\dfrac{a}{58,5}\left(mol\right)\)
So sánh: \(\dfrac{a}{58,5}< \dfrac{a}{53}\Rightarrow\) Dùng `Ca(OH)_2` và `Na_2CO_3` thì điều chế được nhiều `NaOH` hơn
