Question

Answer 1

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

_____0,1--------------->0,1---->0,1

a) m_ZnCl2 = 0,1.136=13,6(g)

b) V_H2 = 0,1.22,4 = 2,24(l)

Answer 2

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

Ta có 1 mol Zn sinh ra 1 mol ZnCl₂

0,1 mol Zn sinh ra 0,1 mol ZnCl₂

\(m_{ZnCl_2}=M.n=136.0,1=13,6\left(g\right)\)

b. 2,24 lít