Đáp án:
\(a,V_{H_2}=4,48\ lít.\\ b,V_{HCl}=0,066364\ lít.\\ c,CM_{ZnCl_2}=3,014M.\)
Giải thích các bước giải:
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2↑\\ n_{Zn}=\dfrac{13}{65}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=n_{Zn}=0,2\ mol.\\ ⇒V_{H_2}=0,2\times 22,4=4,48\ lít.\\ b,Theo\ pt:\ n_{HCl}=2n_{Zn}=0,4\ mol.\\ ⇒m_{HCl}=0,4\times 36,5=14,6\ g.\\ ⇒m_{\text{dd HCl}}=\dfrac{14,6}{20\%}=73\ g.\\ ⇒V_{HCl}=\dfrac{73}{1,1}=66,364\ ml=0,066364\ lít.\\ c,V_{\text{dd\ spư}}=0,066364\ lít.\\ Theo\ pt:\ n_{ZnCl_2}=n_{Zn}=0,2\ mol.\\ ⇒CM_{ZnCl_2}=\dfrac{0,2}{0,066364}=3,014M.\)
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