Đáp án:

 a) 3x10^ 23 phân tử

b)  34g kẽm clorua

c) 14l

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
Zn + 2HCl \to ZnC{l_2} + {H_2}\\
nZn = \dfrac{m}{M} = \dfrac{{16,25}}{{65}} = 0,25\,mol\\
nHCl = 2nZn = 0,5\,mol\\
AHCl = 0,5 \times 6 \times {10^{23}} = 3 \times {10^{23}}\\
b)\\
nZnC{l_2} = nZn = 0,25\,mol\\
mZnC{l_2} = n \times M = 0,25 \times 136 = 34g\\
c)\\
2Zn + {O_2} \to 2ZnO\\
n{O_2} = \dfrac{{0,25}}{2} = 0,125\,mol\\
Vkk = 0,125 \times 5 \times 22,4 = 14l
\end{array}\)

$n_{Zn}=16,25/65=0,25mol$

$PTHH : Zn+2HCl\to ZnCl_2+H_2$

a.Theo pt :

$n_{HCl}=2.n_{Zn}=0,25.2=0,5mol$

$⇒A_{HCl}=0,5.6.10^{23}=3.10^{23}$

b.Theo pt :

$n_{ZnCl_2}=n_{Zn}=0,25mol$

$⇒m_{ZnCl_2}=0,25.136=34g$

Tên muối : Kẽm clorua

$c.2Zn+O_2\overset{t^o}\to 2ZnO$

Theo pt :

$n_{O_2}=1/2.n_{Zn}=1/2.0,25=0,125mol$

$⇒V_{kk}=0,125.22,4.5=14l$