Question

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Answer 1

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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Answer 2

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(a,n_{Al}=n_{AlCl_3}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ m_{Al}=0,2.27=5,4\left(g\right)\\ b,m_{AlCl_3}=133,5.0,2=26,7\left(g\right)\\ c,n_{HCl}=\dfrac{6}{3}.0,3=0,6\left(mol\right)\\ V_{ddHCl}=\dfrac{0,6}{2}=0,3\left(M\right)\)