Đáp án:
\(\begin{array}{l} a,\ V_{H_2}=3,36\ lít.\\ m_{H_2}=0,3\ g.\\ b,\ C\%_{ZnCl_2}=11,99\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:Zn+2HCl\to ZnCl_2+H_2↑\\ n_{Zn}=\dfrac{13}{65}=0,2\ mol.\\ n_{HCl}=0,15\times 2=0,3\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,2}{1}>\dfrac{0,3}{2}\\ ⇒Zn\ \text{dư.}\\ Theo\ pt:\ n_{H_2}=\dfrac12n_{HCl}=0,15\ mol.\\ ⇒V_{H_2}=0,15\times 22,4=3,36\ lít.\\ m_{H_2}=0,15\times 2=0,3\ g.\\ b,\ m_{\text{dd HCl}}=1,05\times 150=157,5\ g.\\ ⇒m_{\text{dd spư}}=m_{Zn}+m_{\text{dd HCl}}-m_{H_2}=13+157,5-0,3=170,2\ g.\\ Theo\ pt:\ n_{ZnCl_2}=\dfrac12n_{HCl}=0,15\ mol.\\ ⇒C\%_{ZnCl_2}=\dfrac{0,15\times 136}{170,2}\times 100\%=11,99\%\end{array}\)
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