Đáp án:
\(\begin{array}{l} b,\ m_{Zn}=1,3\ (g).\\ m_{ZnO}=3,24\ (g).\\ c,\ C_{M_{H_2SO_4\ \text{(dư)}}}=0,9\ (M).\\ C_{M_{ZnSO_4}}=0,6\ (M).\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:\\ Zn+H_2SO_4\to ZnSO_4+H_2\uparrow\ (1)\\ ZnO+H_2SO_4\to ZnSO_4+H_2O\ (2)\\ b,\\ n_{H_2}=\dfrac{0,448}{22,4}=0,02\ (mol).\\ Theo\ pt\ (1):\ n_{Zn}=n_{H_2}=0,02\ (mol).\\ \Rightarrow m_{Zn}=0,02\times 65=1,3\ (g).\\ \Rightarrow m_{ZnO}=4,54-1,3=3,24\ (g).\\ c,\\ \sum n_{H_2SO_4}=0,1\times 1,5=0,15\ (mol).\\ n_{\text{H$_2$SO$_4$ (phản ứng)}}=n_{Zn}+n_{ZnO}=0,02+\dfrac{3,24}{81}=0,06\ (mol)\\ \Rightarrow n_{\text{H$_2$SO$_4$ (dư)}}=0,15-0,06=0,09\ (mol).\\ \sum n_{ZnSO_4}=n_{Zn}+n_{ZnO}=0,02+\dfrac{3,24}{81}=0,06\ (mol).\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=100\ (ml)=0,1\ (l)\\ \Rightarrow C_{M_{H_2SO_4\ \text{(dư)}}}=\dfrac{0,09}{0,1}=0,9\ (M).\\ C_{M_{ZnSO_4}}=\dfrac{0,06}{0,1}=0,6\ (M).\end{array}\)
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