a)
nHCl(bđ) = 0,4.2 = 0,8 (mol)
Zn + 2HCl --> ZnCl2 + H2
Fe + 2HCl --> FeCl2 + H2
=> \(n_{HCl\left(pư\right)}=2.n_{Zn}+2.n_{Fe}=2\left(\dfrac{m_{Zn}}{65}+\dfrac{m_{Fe}}{56}\right)< 2.\left(\dfrac{m_{Zn}+m_{Fe}}{56}\right)=\dfrac{419}{560}< 0,8\)
=> Axit dư
b)
Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 20,95 (1)
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a--->2a------->a----->a
Fe + 2HCl --> FeCl2 + H2
b--->2b------->b----->b
=> a + b= 0,35 (2)
(1)(2) => a = 0,15 (mol); b = 0,2 (mol)
A chứa \(\left\{{}\begin{matrix}ZnCl_2:0,15\left(mol\right)\\FeCl_2:0,2\left(mol\right)\\HCl_{dư}:0,8-0,15.2-0,2.2=0,1\left(mol\right)\end{matrix}\right.\)
\(\left\{{}\begin{matrix}C_{M\left(ZnCl_2\right)}=\dfrac{0,15}{0,4}=0,375M\\C_{M\left(FeCl_2\right)}=\dfrac{0,2}{0,4}=0,5M\\C_{M\left(HCl_{dư}\right)}=\dfrac{0,1}{0,4}=0,25M\end{matrix}\right.\)
