Đáp án:
\(b,\ m_{Zn}=26\ g.\\ c,\ C\%_{ZnSO_4}=25,04\%\)
Giải thích các bước giải:
\(a,\ PTHH:\\ SO_3+H_2O\to H_2SO_4\ (1)\\ Zn+H_2SO_4\to ZnSO_4+H_2↑\ (2)\\ b,\ n_{SO_3}=\dfrac{8,96}{22,4}=0,4\ mol.\\ Theo\ pt\ (1):\ n_{H_2SO_4}=n_{SO_3}=0,4\ mol.\\ Theo\ pt\ (2):\ n_{Zn}=n_{ZnSO_4}=n_{H_2}=n_{H_2SO_4}=0,4\ mol.\\ ⇒m_{Zn}=0,4\times 65=26\ g.\\ c,\ m_{\text{dd H$_2$SO$_4$}}=0,4\times 80+200=232\ g.\\ m_{\text{dd spư}}=m_{Zn}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}\\ ⇒m_{\text{dd spư}}=26+232-0,4\times 2=257,2\ g.\\ ⇒C\%_{ZnSO_4}=\dfrac{0,4\times 161}{257,2}\times 100\%=25,04\%\)
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