Ta có: \(V_{H_2\left(TN_2\right)}>V_{H_2\left(TN_1\right)}\)
=> TN1: Al dư
PTHH:
TN1, TN2: \(2Na+2H_2O\rightarrow2NaOH+H_2\) (1)
\(2NaOH+2Al+2H_2O\rightarrow2NaAlO_2+3H_2\) (2)
TN3: \(2Na+2HCl\rightarrow2NaCl+H_2\) (3)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) (4)
\(Mg+2HCl\rightarrow MgCl_2+H_2\) (5)
- TN1: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT (1), (2): \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{3}{2}n_{NaOH}=\dfrac{1}{2}n_{Na}+\dfrac{3}{2}n_{Na}=2n_{Na}\)
=> \(n_{Na}=\dfrac{0,4}{2}=0,2\left(mol\right)\)
- TN2: \(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
Theo PT (1), (2): \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{3}{2}n_{Al}\)
=> \(n_{Na}=\dfrac{2}{3}.\left(0,7-\dfrac{0,2}{2}\right)=0,4\left(mol\right)\)
- TN3: \(n_{H_2}=\dfrac{26,88}{22,4}=1,2\left(mol\right)\)
Theo PT (3), (4), (5): \(n_{H_2}=\dfrac{1}{2}n_{Na}+\dfrac{3}{2}n_{Al}+n_{Mg}\)
=> \(n_{Mg}=1,2-\dfrac{0,2}{2}-\dfrac{3.0,4}{2}=0,5\left(mol\right)\)
=> m = 0,2.23 + 0,4.27 + 0,5.24 = 27,4 (g)
=> \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,2.23}{27,4}.100\%=16,8\%\\\%m_{Al}=\dfrac{0,4.27}{27,4}.100\%=39,4\%\\\%m_{Mg}=100\%-16,8\%-39,4\%=43,8\%\end{matrix}\right.\)
