\(Đặt:n_{Al}=a\left(mol\right);n_{Al_2O_3}=b\left(mol\right)\left(a,b>0\right)\\a,Al+NaOH+H_2O\rightarrow NaAlO_2+\dfrac{3}{2}H_2\\ Al_2O_3+2NaOH\rightarrow2NaAlO_2+H_2O\\ \Rightarrow\left\{{}\begin{matrix}27a+102b=15,6\\1,5.22,4a=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\\ \Rightarrow n_{NaOH}=n_{NaAlO_2}=a+2b=0,4\left(mol\right)\\ m_{ddNaOH}=\dfrac{0,4.40.100}{5}=320\left(g\right)\\ b,C\%_{ddNaAlO_2}=\dfrac{0,4.82}{15,6+320-0,3.2}.100\approx9,791\%\)
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