Bài 8:
\(m_{HCl}=185,4.100\%=18,54\left(g\right)\)
Gọi x là số mol HCl thêm
\(\Rightarrow\frac{18,54+36,5x}{185,4+36,5x}=0,1657\)
\(\Rightarrow x=0,4\)
\(\Rightarrow V=0,4.22,4=8,96\left(l\right)\)
Bài 4:
a/ Ko tan là Cu
\(m_{AL,Fe}=11,5-3,2=8,3\left(g\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
x_______3x________x _____1,5x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y ___2y_______ y_________y
\(n_{H2}=0,25\left(mol\right)\)
Đặt x, y là mol Al, Fe
\(\left\{{}\begin{matrix}27ax+56y=8,3\\1,5x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
mAlCl3=...........
mFeCl2=............
b/
\(\Sigma n_{HCl}=3x+2y=0,3+0,2=0,5\left(mol\right)\)
\(m_{HCl}=0,5.26,5=18,25\left(g\right)\)
\(m_{dd_{HCl}}=\frac{18,25.100}{14,6}=125\left(g\right)\)
c/ %Al,Cu,Fe=............
d/
\(m_{dd}=11,25+125-3,2-\left(0,25.2\right)=132,55\left(g\right)\)
=>C% AlCl3,FeCl2=.....................
