Tính C% HCl biết 60,5g hỗn hợp Zn và Fe tác dụng với 200g HCl?

Câu 2:

a) \(n_{H_2}=\dfrac{V}{22,4}=\dfrac{12,32}{22,4}=0,55\left(mol\right)\)

\(pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\left(1\right)\\ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }x\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\left(2\right)\\ 2y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }3y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }y\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }3y\)

Từ \(\left(1\right)\) và \(\left(2\right)\), ta có hệ phương trình: \(\left\{{}\begin{matrix}x+3y=0,55\\152x+342y=72,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,25\\y=0,1\end{matrix}\right.\)

\(\Rightarrow m_{Fe}=n\cdot M=0,25\cdot56=14\left(g\right)\\ m_{Al}=n\cdot M=0,1\cdot27=2,7\left(g\right)\)

\(\Rightarrow m_{h^2\left(Al+Fe\right)}=14+2,7=16,7\left(g\right)\)

b) \(n_{H_2SO_4}=x+3y=0,25+3\cdot0,1=0,55\left(mol\right)\)

\(\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{n}{V}=\dfrac{0,3}{0,25}=1,2\left(M\right)\)

\(\text{c) }C_{M\left(FeSO_4\right)}=\dfrac{n}{V}=\dfrac{0,25}{0,25}=1\left(M\right)\\ C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)