a. \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
- Mol theo PTHH : \(1:2:1:1\)
- Mol theo phản ứng : \(0,1\rightarrow0,2\rightarrow0,1\rightarrow0,1\)
\(\Rightarrow m_{HCl}=n_{HCl}.M_{HCl}=0,1.\left(2+35,5\right)=3,75\left(g\right)\)
b. Từ a. suy ra : \(m_{ZnCl_2}=n_{ZnCl_2}.M_{ZnCl_2}=0,2.\left(65+71\right)=27,2\left(g\right)\)
c. Từ a. suy ra : \(V_{H_2}=n_{H_2}.22,4=0,1.22,4=2,24\left(l\right)\)
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