- Xét phần 1: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Na}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2Na+2HCl\rightarrow2NaCl+H_2\)
a---------------------------->0,5a
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
b----------------------------->b
`=> 0,5a + b = 0,05 (1)`
- Xét phần 2: Đặt hệ số tỉ lệ \(\dfrac{P_2}{P_1}=k\left(k>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=ak\left(mol\right)\\n_{Mg}=bk\left(mol\right)\end{matrix}\right.\)
Ta có: \(m_{gi\text{ả}m}=m_{O\left(CuO\right)\left(p\text{ư}\right)}=8-6,72=1,28\left(g\right)\)
\(\Rightarrow n_{CuO\left(p\text{ư}\right)}=n_{O\left(p\text{ư}\right)}=\dfrac{1,28}{16}=0,08\left(mol\right)\)
PTHH:
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
ak-------------------------->0,5ak
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,08-->0,08
`=> 0,5ak = 0,08 <=> ak = 0,16 (2)`
- Xét hỗn hợp ban đầu:
Ta có: \(\dfrac{hhb\text{đ}}{P_1}=\dfrac{P_1+P_2}{P_1}=\dfrac{P_1+k.P_1}{P_1}=k+1\)
\(\Rightarrow\left\{{}\begin{matrix}n_{Na}=a\left(k+1\right)\left(mol\right)\\n_{Mg}=b\left(k+1\right)\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow23a\left(k+1\right)+24b\left(k+1\right)=8,2\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow\left\{{}\begin{matrix}a=0,04\\b=0,03\\k=4\end{matrix}\right.\left(TM\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Na}=\dfrac{0,04.23}{0,04.23+0,03.24}.100\%=56,1\%\\\%m_{Mg}=100\%-56,1\%=43,9\%\end{matrix}\right.\)
