a, Ta có: \(n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
___0,3_____0,9____0,3____0,45 (mol)
\(m_{Al}=0,3.27=8,1\left(g\right)\)
\(\Rightarrow\%m_{Cu}=\dfrac{20-8,1}{20}.100\%=59,5\%\)
b, Có: \(m_{ddHCl}=\dfrac{0,9.36,5}{10\%}=328,5\left(g\right)\)
m dd sau pư = 8,1 + 328,5 - 0,45.2 = 335,7 (g)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,3.133,5}{335,7}.100\%\approx11,93\%\)
Bạn tham khảo nhé!