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Giải thích các bước giải:

\(3/\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2↑\\ n_{Fe}=\dfrac{22,4}{56}=0,4\ mol.\\ n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,4}{1}>\dfrac{0,25}{1}\\ ⇒Fe\ dư.\\ Theo\ pt:\ n_{FeSO_4}=n_{H_2}=n_{H_2SO_4}=0,25\ mol.\\ ⇒m_{FeSO_4}=0,25.152=38\ g.\\ ⇒V_{H_2}=0,25.22,4=5,6\ lít.\\ 4/\\ PTHH:CuO+H_2SO_4\to CuSO_4+H_2O\\ n_{CuO}=\dfrac{1,6}{80}=0,02\ mol.\\ n_{H_2SO_4}=\dfrac{100\times 20\%}{98}≈0,2\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,02}{1}<\dfrac{0,2}{1}\\ ⇒H_2SO_4\ dư.\\ ⇒n_{H_2SO_4}(dư)=0,2-0,02=0,18\ mol.\\ m_{\text{dd spư}}=n_{CuO}+m_{\text{dd H$_2$SO$_4$}}=1,6+100=101,6\ g.\\ Theo\ pt:\ n_{CuSO_4}=n_{CuO}=0,02\ mol.\\ ⇒C\%_{CuSO_4}=\dfrac{0,02\times 160}{101,6}\times 100\%=3,15\%\\ C\%_{H_2SO_4}(dư)=\dfrac{0,18\times 98}{101,6}\times 100\%=17,36\%\\ 5/\\ PTHH:CO_2+2NaOH\to Na_2CO_3+H_2O\\ n_{CO_2}=\dfrac{1,568}{22,4}=0,07\ mol.\\ n_{NaOH}=\dfrac{6,4}{40}=0,16\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,07}{1}<\dfrac{0,16}{2}\\ \Rightarrow NaOH\ dư.\\ Theo\ pt:\ n_{Na_2CO_3}=n_{CO_2}=0,07\ mol.\\ \Rightarrow m_{Na_2CO_3}=0,07.106=7,42\ g.\\ 6/\\ a,PTHH:CaCl_2+2AgNO_3\to Ca(NO_3)_2+2AgCl↓\\ n_{CaCl_2}=\dfrac{2,22}{111}=0,02\ mol.\\ n_{AgNO_3}=\dfrac{1,7}{170}=0,01\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,02}{1}>\dfrac{0,01}{2}\\ \Rightarrow CaCl_2\ dư.\\ Theo\ pt:\ n_{AgCl}=n_{AgNO_3}=0,01\ mol.\\ \Rightarrow m_{AgCl}=0,01.143,5=1,435\ g.\\ b,Theo\ pt:\ n_{Ca(NO_3)_2}=\dfrac{1}{2}n_{AgNO_3}=0,005\ mol.\\ n_{CaCl_2}(dư)=0,02-\dfrac{0,01}{2}=0,015\ mol.\\ V_{dung\ dịch\ spư}=30+70=100\ ml=0,1\ lít.\\ ⇒CM_{Ca(NO_3)_2}=\dfrac{0,005}{0,1}=0,05M.\\ \Rightarrow CM_{CaCl_2}(dư)=\dfrac{0,015}{0,1}=0,15M.\\ 7/\\ a,PTHH:CuCl_2+2NaOH\to Cu(OH)_2+2NaCl\\ n_{NaOH}=\dfrac{20}{40}=0,5\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,2}{1}<\dfrac{0,5}{2}\\ \Rightarrow NaOH\ dư.\\ Theo\ pt:\ n_{Cu(OH)_2}=n_{CuCl_2}=0,2\ mol.\\ \Rightarrow m_{Cu(OH)_2}=0,2.98=19,6\ g.\\ b,\ m_{NaOH}(dư)=0,5-(0,2.2)=0,1 mol.\\ \Rightarrow m_{NaOH}(dư)=0,1.40=4\ g.\\ Theo\ pt:\ n_{NaCl}=2n_{CuCl_2}=0,4\ mol.\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4\ g.\\ 8/\\ PTHH:Fe+CuSO_4\to FeSO_4+Cu\\ m_{\text{dd CuSO$_4$}}=100\times 1,12=112\ g.\\ \Rightarrow n_{CuSO_4}=\dfrac{112\times 10\%}{160}=0,07\ mol.\\ n_{Fe}=\dfrac{1,96}{56}=0,035\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,035}{1}<\dfrac{0,07}{1}\\ \Rightarrow CuSO_4\ dư.\\ \Rightarrow n_{CuSO_4}(dư)=0,07-0,035=0,035\ mol.\\ Theo\ pt:\ n_{FeSO_4}=n_{Cu}=n_{Fe}=0,035\ mol.\\ \text{Đổi 100 ml = 0,1 lít.}\\ \Rightarrow CM_{FeSO_4}=\dfrac{0,035}{0,1}=0,35M.\\ CM_{Cu}=\dfrac{0,035}{0,1}=0,35M.\\ CM_{CuSO_4}(dư)=\dfrac{0,035}{0,1}=0,35M.\)

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