a) Gọi số mol Mg, Fe là a, b (mol)
=> 24a + 56b = 8 (1)
\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
a---->2a------>a------>a
Fe + 2HCl --> FeCl2 + H2
b--->2b------>b---->b
=> 2a + 2b = 0,4 (2)
(1)(2) => a = 0,1 (mol); b = 0,1 (mol)
=> \(\left\{{}\begin{matrix}m_{Mg}=0,1.24=2,4\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)
b)
\(\left\{{}\begin{matrix}m_{MgCl_2}=0,1.95=9,5\left(g\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\end{matrix}\right.\)
\(n_{H_2}=a+b=0,2\left(mol\right)\)
=> \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
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