a)
Gọi số mol Zn, Fe là a, b (mol)
=> 65a + 56b = 43,7 (1)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
a--->2a------------->a
Fe + 2HCl --> FeCl2 + H2
b--->2b-------------->b
=> a + b = 0,7 (2)
(1)(2) => a = 0,5 (mol); b = 0,2 (mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,5.65=32,5\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
b) nHCl = 2a + 2b = 1,4 (mol)
\(\Rightarrow V_{dd.HCl}=\dfrac{1,4}{0,05}=28\left(l\right)\)
c)
\(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)
PTHH: Fe3O4 + 4H2 --to-> 3Fe + 4H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,7}{4}\) => Fe3O4 dư, H2 hết
PTHH: Fe3O4 + 4H2 --to-> 3Fe + 4H2O
0,7----->0,525
=> mFe = 0,525.56 = 29,4 (g)