Question

Answer 1

a) \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b) \(n_{Zn}=\dfrac{16}{65}\left(mol\right)\)

Theo PTHH: \(n_{H_2}=n_{Zn}=\dfrac{16}{65}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{16}{65}.22,4=\dfrac{1792}{325}\left(l\right)\)

c) Theo PTHH: \(n_{HCl}=2.n_{Zn}=\dfrac{32}{65}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{32}{65}.36,5=\dfrac{1168}{65}\left(g\right)\)

Answer 2

a)

$Zn + 2HCl \to ZnCl_2 + H_2$
b)

Theo PTHH : $n_{H_2} = n_{Zn} = \dfrac{16}{65} (mol)$
$V_{H_2} = \dfrac{16}{65}.22,4 = 5,51(lít)$

c)

$n_{HCl} = 2n_{Zn} = \dfrac{32}{65}(mol)$
$m_{HCl} = \dfrac{32}{65}.36,5 = 17,97(gam)$