Đáp án:
$b,V_{H_2}=2,24l.$
$c,C\%_{ddZnCl_2}=16,5\%$
Giải thích các bước giải:
$a,PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$
$b,n_{Zn}=\dfrac{6,5}{65}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,1mol.$
$⇒V_{H_2}=0,1.22,4=2,24l.$
$c,Theo$ $pt:$ $n_{HCl}=2n_{Zn}=0,2mol.$
$⇒m_{HCl}=0,2.36,5=7,3g.$
$⇒m_{ddHCl}=\dfrac{7,3}{9,6\%}=76,04g.$
$Theo$ $pt:$ $n_{ZnCl_2}=n_{Zn}=0,1mol.$
$⇒m_{ZnCl_2}=0,1.136=13,6g.$
$⇒m_{H_2}=0,1.2=0,2mol.$
$⇒m_{dd sau pư}=76,04+6,5-0,2=82,34g.$
$⇒C\%_{ddZnCl_2}=\dfrac{13,6}{82,34}.100\%=16,5\%$
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