a)
\(n_{HCl\left(bđ\right)}=\dfrac{65,7}{36,5}=1,8\left(mol\right)\)
- Giả sử axit dư
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
Theo PTHH:
\(n_{HCl\left(pư\right)}=2\left(n_{Fe}+n_{Mg}\right)=2\left(\dfrac{m_{Fe}}{56}+\dfrac{m_{Mg}}{24}\right)< 2.\dfrac{20}{24}=\dfrac{5}{3}< 1,8\left(mol\right)\)
=> Giả sử đúng
b)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
Gọi số mol Fe, Mg là a, b (mol)
=> 56a + 24b = 20 (1)
Theo PTHH: \(n_{H_2}=a+b=0,7\left(mol\right)\) (2)
(1)(2) => a = 0,1 (mol); b = 0,6 (mol)
\(\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{Mg}=0,6.24=14,4\left(g\right)\end{matrix}\right.\)
