Câu 6 :
\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.4.........1.2.......0.4..............0.6\)
\(m_{dd_{HCl}}=\dfrac{1.2\cdot36.5\cdot100}{10.95}=400\left(g\right)\)
\(m_{dd}=10.8+400-0.6\cdot2=409.6\left(g\right)\)
\(C\%AlCl_3=\dfrac{0.4\cdot133.5}{409.6}\cdot100\%=13.04\%\)
Câu 7 :
\(n_{Al}=\dfrac{10.8}{27}=0.4\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.4..........1.2..........0.4............0.6\)
\(V_{dd_{HCl}}=\dfrac{1.2}{2}=0.6\left(l\right)\)
\(C_{M_{AlCl_3}}=\dfrac{0.4}{0.6}=0.67\left(M\right)\)
