a/ $Zn(OH)_2+2HCl\to ZnCl_2+2H_2O$

$n_{Zn(OH)_2}=\dfrac{19,8}{99}=0,2(mol)$

Theo PTHH: $n_{HCl}=2n_{Zn(OH)_2}$

$→n_{HCl}=0,4(mol)\\→m_{HCl}=0,4.36,5=14,6g\\C\%_{HCl}=\dfrac{14,6}{146}=10\%$

Vậy $C\%_{HCl}=10\%$

b/ Theo PTHH: $n_{ZnCl_2}=n_{Zn(OH)_2}$

$→n_{ZnCl_2}=0,2(mol)\\→m_{ZnCl_2}=0,2.136=27,2g$

Vậy $m_{ZnCl_2}=27,2g$

c/ Không có khí sinh ra

Bạn tham khảo!

$\text{Ta có PTHH sau:}$ 

$\text{Zn(OH)2+2HCl$\rightarrow$ZnCl2+2H2O}$

$\text{nZn(OH)2=$\frac{m}{M}$=$\frac{19,8}{99}$=0,2 mol}$

$\rightarrow$ $\text{nHCl=2nZn(OH)2=0,2.2=0,4 mol}$

$\text{mHCl=n.M=0,4.36,5=14,6g}$

$\text{a) C% HCl=$\frac{mct}{mdd}$.100$\frac{14,6}{146}$.100=10%}$

$\text{b) nZnCl2=nZn(OH)2=0,2 mol}$

$\rightarrow$ $\text{mZnCl2=n.M=0,2.136=27,2g}$

$\text{c) Vì sau phản ứng không có khí sinh ra nên không tính được}$

#TRANPHAMGIABAO