Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\\n_{Cu}=c\left(mol\right)\end{matrix}\right.\)
=> 27a + 65b + 64c = 20,1 (1)
\(n_{H_2}=\dfrac{6,048}{22,4}=0,27\left(mol\right)\)
PTHH:
`2Al + 6HCl -> 2AlCl_3 + 3H_2`
a----------------------------->1,5a
`Zn + 2HCl -> ZnCl_2 + H_2`
b----------------------------->b
=> 1,5a + b = 0,27 (2)
\(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)
a---------------->0,5a
\(2Zn+O_2\xrightarrow[]{t^o}2ZnO\)
b--------------->b
\(2Cu+O_2\xrightarrow[]{t^o}2CuO\)
c----------------->c
=> 0,5a.102 + 81b + 80c = 26,82 (3)
(1);(2);(3) => a = 0,1; b = 0,12; c = 0,15
=> m = mCu = 0,15.64 = 9,6 (g)
=> \(\left\{{}\begin{matrix}\%n_{Al}=\dfrac{0,1}{0,1+0,12+0,15}.100\%=27,027\%\\\%n_{Zn}=\dfrac{0,12}{0,1+0,12+0,15}.100\%=32,342\%\\\%n_{Cu}=\left(100-27,027-32,342\right)\%=40,631\%\end{matrix}\right.\)
Gọi $n_{Al} = a(mol) ; n_{Zn} = b(mol) ; n_{Cu} = c(mol) \Rightarrow 27a + 65b + 64c = 20,1(1)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
Theo PTHH : $n_{H_2} = 1,5a + b = 0,27(mol)(2)$
$4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3$
$2Zn + O_2 \xrightarrow{t^o} 2ZnO$
$2Cu + O_2 \xrightarrow{t^o} 2CuO$
$m_{oxit} = m_{Al_2O_3} + m_{ZnO} + m_{CuO} = 0,5a.102 + 81b + 80c = 26,82(3)$
Từ (1)(2)(3) suy ra a = 0,1 ; b = 0,12 ; c = 0,15
$\Rightarrow m = m_{Cu} = 0,15.64 = 9,6(gam)$
$\%n_{Al} = \dfrac{0,1}{0,1 + 0,12 + 0,15}.100\% = 27,02\%$
$\%n_{Zn} = \dfrac{0,12}{0,1 + 0,12 + 0,15}.100\% = 32,43\%$
$\%n_{Cu} = \dfrac{0,15}{0,1 + 0,12 + 0,15}.100\% = 40,55\%$
