\(2Na+2H2O-->2NaOH+H2\)
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(n_{Na}=2n_{H2}=0,4\left(mol\right)\)
\(m_{Na}=0,4.23=9,2\left(g\right)\)
\(n_{Al}=2n_{Na}=0,8\left(mol\right)\)
\(m_{Al}=0,8.27=21,6\left(g\right)\)
\(\%m_{Al}=\frac{21,6}{21,6+9,2}.100\%=69,23\%\)
Giả sử số mol Na là x suy ra Al là 2x.
Phản ứng xảy ra:
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(\Rightarrow n_{NaOH}=n_{Na}=x\left(mol\right)\)
\(NaOH+Al+H_2O\rightarrow NaAlO_2+\frac{3}{2}H_2\)
Vì : \(n_{Na}>n_{NaOH}\) nên Al dư.
\(\Rightarrow n_{H2}=\frac{1}{2}n_{Na}+\frac{3}{2}n_{NaOH}=\frac{1}{2}x+\frac{3}{2}x=2x=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow x=0,1\)
\(\Rightarrow m=m_{Na}+m_{Al}=23x+27.2x=7,7\left(g\right)\)
\(\Rightarrow m_{Al}=27.2x=5,4\)
\(\Rightarrow\%m_{Al}=\frac{5,4}{7,7}=70,13\%\)
