\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.2.........0.6...........0.2\)
\(m_{AlCl_3}=0.2\cdot133.5=26.7\left(g\right)\)
\(V_{dd_{HCl}}=\dfrac{0.6}{3}=0.2\left(l\right)\)
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