\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right);n_{NaOH}=\dfrac{112.0,5.20\%}{40}=0,28\left(mol\right)\\ Ta.có:1< T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,28}{0,15}\approx1,867< 2\\ \Rightarrow Sp:Na_2CO_3,NaHCO_3\\ CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\left(1\right)\\ NaOH+CO_2\rightarrow NaHCO_3\left(2\right)\\ Đặt:a=n_{CO_2\left(1\right)};b=n_{CO_2\left(2\right)}\left(a,b>0\right)\\ \Rightarrow\left\{{}\begin{matrix}a+b=0,15\\2a+b=0,28\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,13\\b=0,02\end{matrix}\right.\\ \Rightarrow m_{muối}=m_{Na_2CO_3}+m_{NaHCO_3}=106a+84b=15,46\left(g\right)\)
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