Question

Answer 1

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

        0,2-->0,4--------------->0,2

=> \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

c) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

Answer 2

\(n_{Zn}=\dfrac{13}{65}=0,2mol\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,2      0,4                        0,2

\(m_{HCl}=0,4\cdot36,5=14,6g\)

\(V_{H_2}=0,2\cdot22,4=4,48l\)