Đáp án:
$b,m_{ddHCl}=100g.$
$c,C\%_{ZnCl_2}=12,58\%$
Giải thích các bước giải:
$a,PTPƯ:ZnO+2HCl\xrightarrow{} ZnCl_2+H_2O$
$b,n_{ZnO}=\dfrac{8,1}{81}=0,1mol.$
$Theo$ $pt:$ $n_{HCl}=2n_{ZnO}=0,2mol.$
$⇒m_{HCl}=0,2.36,5=7,3g.$
$⇒m_{ddHCl}=\dfrac{7,3}{7,3\%}=100g.$
$c,m_{ddspư}=m_{ZnO}+m_{ddHCl}=8,1+100=108,1g.$
$Theo$ $pt:$ $n_{ZnCl_2}=n_{ZnO}=0,1mol.$
$⇒m_{ZnCl_2}=0,1.136=13,6g.$
$⇒C\%_{ZnCl_2}=\dfrac{13,6}{108,1}.100\%=12,58\%$
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