\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.1....................0.1.........0.1\)
\(m_{ZnCl_2}=0.1\cdot136=13.6\left(g\right)\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
- n\(_{Zn}\)=0,1 (mol)
pư: Zn+2HCl --> ZnCl\(_2\)+H\(_2\)
0,1 0,2 0,1 0,1 (mol)
=> m\(ctHCl\)=7,3 (g)
m\(_{ctZnCl_2}\)=13,6 (g)
m\(_{H_2}\)=0,2 (g) ; V\(_{H_2}\)=2,24 (l)
=> C%\(_{HCl}\)=\(\dfrac{7,3}{70}\)x100%\(\approx\)10,43%
Chẳng biết bạn hỏi gì nên cứ tính đại vậy :)))
