PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ 0,1mol:0,1mol\leftarrow0,1mol:0,1mol\)
\(Cu+H_2SO_4\rightarrow X\) (Cu không PƯ với \(H_2SO_4\) )
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(m_{Zn}=0,1.65=6,5\left(g\right)\)
\(\Leftrightarrow m_{Cu}=10,5-6,5=4\left(g\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{6,5}{10,5}100\%=61,9\%\\\%m_{Cu}=100\%-61,9\%=38,1\%\end{matrix}\right.\)
Ta có
PT \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) (1)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo (1) ta có \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Zn}=65\times0,1=6,5\left(g\right)\)
\(\Rightarrow\%Zn=\dfrac{6,5}{10,5}\times100\%\approx61,9\%\)
\(\Rightarrow m_{Cu}=10,5-6,5=4\left(g\right)\)\(\Rightarrow\%Cu=\dfrac{4}{10,5}\times100\%\approx38,1\%\)
nH2 = \(\dfrac{2,24}{22,4}\) = 0,1mol
- cho hỗn hợp 2 KL vào dd H2SO4 loãng chỉ có Zn phản ứng
Zn + H2SO4 -> ZnSO4 + H2 \(\uparrow\)
0,1<---------------------0,1
%Zn = \(\dfrac{0,1.65}{10,5}.100\%\) \(\approx\)61,9%
=>%Cu = 100% - 61,9% = 38,1%
