Đáp án:
$b,m=m_{Zn}=3,62g.$
$c,V_{H_2}=1,12l.$
$d,m_{ZnCl_2}=6,8g.$
Giải thích các bước giải:
$a,PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$
$b,m_{HCl}=50.7,3\%=3,65g.$
$⇒n_{HCl}=\dfrac{3,65}{36,5}=0,1mol.$
$Theo$ $pt:$ $n_{Zn}=\dfrac{1}{2}n_{HCl}=0,05mol.$
$⇒m=m_{Zn}=0,05.65=3,25g.$
$c,Theo$ $pt:$ $n_{H_2}=\dfrac{1}{2}n_{HCl}=0,05mol.$
$⇒V_{H_2}=0,05.22,4=1,12l.$
$d,Theo$ $pt:$ $n_{ZnCl_2}=\dfrac{1}{2}n_{HCl}=0,05mol.$
$⇒m_{ZnCl_2}=0,05.136=6,8g.$
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