Đáp án:

$a,V_{H_2}=6,72l.$

$b,C\%_{ZnCl_2}=34,17\%$

$⇒C\%_{HCl}(dư)=12.23\%$

Giải thích các bước giải:

$a,PTPƯ:Zn+2HCl→ZnCl_2+H_2↑$

$n_{Zn}=\dfrac{20}{65}=0,3mol.$

$m_{HC}=100.36,5\%=36,5g.$

$⇒n_{HCl}=\dfrac{36,5}{36,5}=1mol.$

$\text{Lập tỉ lệ:}$

$\dfrac{0,3}{1}<\dfrac{1}{2}$

$⇒n_{HCl}$ $dư.$

$\text{⇒Tính theo}$ $n_{Zn}$

$Theo$ $pt:$ $n_{H_2}=n_{Zn}=0,3mol.$

$⇒V_{H_2}=0,3.22,4=6,72l.$

$b,Theo$ $pt:$ $n_{ZnCl_2}=n_{Zn}=0,3mol.$

$⇒m_{ZnCl_2}=0,3.136=40,8g.$

$⇒m_{H_2}=0,3.2=0,6g.$

$⇒n_{HCl}(dư)=1-\dfrac{0,3.2}{1}=0,4mol.$

$⇒m_{HCl}(dư)=0,4.36,5=14,6g.$

$⇒m_{ddZnCl_2}=20+100-0,6=119,4g.$

$⇒C\%_{ZnCl_2}=\dfrac{40,8}{119,4}.100\%=34,17\%$

$⇒C\%_{HCl}(dư)=\dfrac{14,6}{119,4}.100\%=12.23\%$

chúc bạn học tốt!

$n_{Zn}=20/65=0,31mol$

$m_{HCl}=100.36,5\%=36,5g$

$⇒n_{HCl}=36,5/36,5=1mol$

$PTHH :$

                      $Zn    +    2HCl\to ZnCl_2+H_2↑$

$\text{a/Theo pt :    1 mol         2 mol}$

$\text{Theo đbài :    0,31mol       1 mol}$

$\text{⇒Sau pư HCl dư  }$

$⇒n_{H_2}=n_{Zn}=0,31mol$

$⇒V_{H_2}=0,31.22,4=6,944l$

$\text{c/Theo pt :}$

$n_{ZnCl_2}=n_{Zn}=0,31mol$

$⇒m_{ZnCl_2}=0,31.136=42,16g$

$m_{H_2}=0,31.2=0,62g$

$n_{HCl pư}=2.n_{Zn}=2.0,31=0,62mol$

$⇒n_{HCl dư}=1-0,62=0,38mol$

$⇒m_{HCl dư}=0,38.36,5=13,87g$

$⇒m_{\text{dd ZnCl_2}}=20+100-0,62=119,38g$

$⇒C\%_{ZnCl_2}=\dfrac{42,16}{119,38}.100\%=35,31\%$

$C\%_{HCl dư}=\dfrac{13,87}{119,4}.100\%=11,61\%$