200ml = 0,2l
\(n_{HCl}=0,15.0,2=0,03\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,015 0,03 0,015 0,015
a) \(n_{Fe}=\dfrac{0,03.1}{2}=0,015\left(mol\right)\)
⇒ \(m_{Fe}=0,015.56=0,84\left(g\right)\)
b) \(n_{H2}=\dfrac{0,03.1}{2}=0,015\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,015.22,4=0,336\left(l\right)\)
c) \(n_{ZnCl2}=\dfrac{0,015.1}{1}=0,015\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,015}{0,2}=0,075\left(M\right)\)
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