Mg + 2HCl → MgCl2 + H2 (1)
2Al + 6HCl → 2AlCl3 + 3H2 (2)
Cu + HCl → X
a) Chất rắn không tan là Cu
\(\Rightarrow\%Cu=\dfrac{6,4}{11,5}\times100\%=55,65\%\)
\(\Rightarrow m_{Fe}+m_{Al}=11,5-6,4=5,1\left(g\right)\)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,25\times2=0,5\left(g\right)\)
Gọi \(x,y\) lần lượt là số mol của Mg và Al
Theo PT1: \(n_{H_2}=n_{Mg}=x\left(mol\right)\)
Theo PT2: \(n_{H_2}=\dfrac{3}{2}n_{Al}=1,5y\left(mol\right)\)
Ta có: \(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
Vậy \(n_{Mg}=0,1\left(mol\right);n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,1\times24=2,4\left(g\right)\)
\(\Rightarrow\%Mg=\dfrac{2,4}{11,5}\times100\%=20,87\%\)
\(m_{Al}=0,1\times27=2,7\left(g\right)\)
\(\Rightarrow\%Al=\dfrac{2,7}{11,5}\times100\%=23,48\%\)
b) Theo PT1: \(n_{HCl}=2n_{Mg}=2\times0,1=0,2\left(mol\right)\)
Theo PT2: \(n_{HCl}=3n_{Al}=3\times0,1=0,3\left(mol\right)\)
\(\Rightarrow\Sigma n_{HCl}=0,2+0,3=0,5\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,5\times36,5=18,25\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{18,25}{10\%}=182,5\left(g\right)\)
\(\Rightarrow m_{dd}saupư=m_{hhX}+m_{ddHCl}-m_{H_2}=11,5+182,5-0,5=193,5\left(g\right)\)
Theo PT1: \(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)
\(\Rightarrow m_{MgCl_2}=0,1\times95=9,5\left(g\right)\)
\(\Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{193,5}\times100\%=4,91\%\)
Theo PT2: \(n_{AlCl_3}=n_{Al}=0,1\left(mol\right)\)
\(\Rightarrow m_{AlCl_3}=0,1\times133,5=13,35\left(g\right)\)
\(\Rightarrow C\%_{AlCl_3}=\dfrac{13,35}{193,5}\times100\%=6,89\%\)
BÀI 1: Cho 4,93g hỗn hợp A gồm Mg, Zn vào 250g ddHCl 7,3%
