Question

Answer 1

nZn=6,5/65=0,1 mol

Zn + 2HCl --> ZnCl2 + H2

0,1     0,2                      0,1             mol

CM HCl = 0,2/0,1=2M

VH2 = 0,1 *22,4=2,24 lít

Answer 2

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,2\left(mol\right)\\n_{H_2}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)