a: \(n_{Zn}=\dfrac{18.2}{65}=0.28\left(mol\right)=n_{ZnCl_2}\)
\(\Leftrightarrow n_{HCl}=0.56\left(mol\right)\)
\(m_{HCl}=0.56\cdot36.5=20.44\left(g\right)\)
b: \(n_{H_2}=0.28\left(mol\right)\)
\(V_{H_2}=0.28\cdot22.4=6.272\left(lít\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
a. \(n_{Zn}=\dfrac{18,2}{65}=0,28\left(mol\right)\)
Theo PTHH ta có: \(n_{HCl}=n_{Zn}\cdot\dfrac{2}{1}=0,28\cdot\dfrac{2}{1}=0,56\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,56.\left(1+35,5\right)=20,44\left(g\right)\)
b. Theo PTHH ta có: \(n_{H_2}=n_{Zn}\cdot\dfrac{1}{1}=0,28\cdot\dfrac{1}{1}=0,28\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,28.22,4=6,272\left(l\right)\)
c. Cách 1:
- Theo PTHH ta có: \(n_{ZnCl_2}=n_{Zn}\cdot\dfrac{1}{1}=0,28\cdot\dfrac{1}{1}=0,28\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=0,28.\left(65+35,5.2\right)=38,08\left(g\right)\)
Cách 2:
- \(m_{H_2}=0,28.1.2=0,56\left(g\right)\)
- Theo định luật bảo toàn khối lượng ta có:
\(m_{Zn}+m_{HCl}=m_{ZnCl_2}+m_{H_2}\)
\(\Rightarrow m_{ZnCl_2}=m_{Zn}+m_{HCl}-m_{H_2}=18,2+20,44-0,56=38,08\left(g\right)\)
