Đáp án:
\(a,\ m_1=2,6\ g.\\ m_2=20\ g.\\ b,\ C\%_{ZnCl_2}=24,16\%\)
Giải thích các bước giải:
\(a,\ PTHH:Zn+2HCl\to ZnCl_2+H_2↑\\ n_{H_2}=\dfrac{0,896}{22,4}=0,04\ mol.\\ Theo\ pt:\ n_{Zn}=n_{H_2}=0,04\ mol.\\ ⇒m_1=m_{Zn}=0,04\times 65=2,6\ g.\\ Theo\ pt:\ n_{HCl}=2n_{H_2}=0,08\ mol.\\ ⇒m_2=m_{\text{dd HCl}}=\dfrac{0,08\times 36,5}{14,6\%}=20\ g.\\ b,\\\ m_{\text{dd spư}}=m_{Zn}+m_{\text{dd HCl}}-m_{H_2}\\\,\,\,\,\,\,\,\,\,=2,6+20-0,04\times 2=22,52\ g.\\ Theo\ pt:\ n_{ZnCl_2}=n_{H_2}=0,04\ mol.\\ ⇒C\%_{ZnCl_2}=\dfrac{0,04\times 136}{22,52}\times 100\%=24,16\%\)
chúc bạn học tốt!