a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,2 0,1 0,1
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c, \(C_{M_{ddHCl}}=\dfrac{0,2}{0,25}=0,8M\)
Bài 1 :
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,1 0,2 0,1 0,1
a) \(n_{ZnCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,1.136=13,6\left(g\right)\)
b) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
c) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
250ml = 0,25l
\(C_{M_{ddHCl}}=\dfrac{0,2}{0,25}=0,8\left(M\right)\)
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