\(a.\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(b.\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
\(a.........................a\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(\dfrac{2}{3}a........................a\)
\(m_{Mg}=24a\left(g\right)\)
\(m_{Al}=\dfrac{2}{3}a\cdot27=18a\left(g\right)\)
\(m_{Al}< m_{Mg}\)
=> Cần dùng Al là nhỏ nhất
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\(2.\)
\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{t^0}}2Fe+3H_2O\)
\(ZnO+H_2\underrightarrow{^{t^0}}Zn+H_2O\)
\(n_{H_2O}=n_{H_2}=\dfrac{5.4}{18}=0.3\left(mol\right)\)
\(BTKL:\)
\(m_A+m_{H_2}=m_{kl}+m_{H_2O}\)
\(\Rightarrow m_{kl}=20+0.3\cdot2-5.4=15.2\left(g\right)\)
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