a,
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
Gọi số mol Zn,Al lần lượt a;ba;b
PTHH:
\(2Zn+O_2\rightarrow2ZnO\)
\(4Al+3O_2\rightarrow Al_2O_3\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}65a+27b=14,4\\0,5a+0,75b=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=\frac{27}{235}\left(mol\right)\\b=\frac{181}{705}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Zn}=\frac{27}{235}.65=\frac{351}{47}\left(g\right)\)
\(\Rightarrow\%m_{Zn}=\frac{\frac{351}{47}}{14,4}.100\%=51,86\%\)
\(\Rightarrow\%m_{Al}=100\%-51,86\%=48,14\%\)
b, Trong 14,4g hỗn hợp Zn;Al chứa \(\frac{27}{235}\left(mol\right)\) Zn và \(\frac{181}{705}\left(mol\right)\) Al
\(\Rightarrow\) Trong 9,2g hỗn hợp Zn;Al chứa \(\frac{69}{940}\left(mol\right)\) Zn và \(\frac{4163}{25380}\left(mol\right)\) Al
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
Ta có :
\(n_{H2}=n_{Zn}+\frac{1}{5}n_{Al}=\frac{69}{940}+1,5.\frac{4163}{25380}=\frac{23}{72}\left(mol\right)\)
\(\Rightarrow V_{H2}=\frac{23}{72}.22,4=7,16\left(l\right)\)
a) Đặt \(n_{Zn}=x;n_{Al}=y\left(mol\right)\)
\(\Rightarrow65x+27y=14,4\left(1\right)\)
\(n_{O_2}=0,25\left(mol\right)\)
QT nhường e: Zn ----> Zn+2 + 2e
____________x_____________2x
Al ----> Al+3 + 3e
y______________3y
QT nhận e: O2 + 4e ---> 2O-2
__________0,25___1
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}x=\\y=\end{matrix}\right.\)