Sửa A. Các loại vật liệu tự nhiên

A

Bài 5
CuO<Cu(OH)₂<CuCl₂<CuSO₄
Bài 6
\(\%R=\dfrac{2R}{2R+48}\cdot100\%=52,94\%\\ \Rightarrow R=27,Al\)
Bài 7
\(CTPT\left(Z\right):H_xS_y\\ M_Z=17.2=34g/mol=x+32y\\ \%H=\dfrac{x}{x+32y}\cdot100\%=5,88\%\\ \Leftrightarrow\dfrac{x}{34}\cdot100\%=5,88\%\\ \Rightarrow x\approx2\\ y=\dfrac{34-2}{32}=1\\ \Rightarrow CTPT\left(X\right):H_2S\)

\(a,Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\\ b,Fe_2O_3+SO_3\rightarrow Không.p.ứ\\ c,Na_2O+H_2O\rightarrow2NaOH\\ d,Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\\ e,SO_3+H_2O\rightarrow H_2SO_4\\ f,2NaOH\left(dư\right)+CO_2\rightarrow Na_2CO_3+H_2O\\ g,P_2O_5+3H_2O\rightarrow2H_3PO_4\\ h,CaO+CO_2\rightarrow CaCO_3\)

\(n_{Fe}=\dfrac{3,92}{56}=0,07\left(mol\right)\\ a,Fe+CuSO_4\rightarrow FeSO_4+Cu\\ b,n_{Cu}=n_{FeSO_4}=n_{Fe}=0,07\left(mol\right)\\ b,m_{FeSO_4}+m_{Cu}=0,07.\left(152+64\right)=0,07.216=15,12\left(g\right)\\ c,C_{MddCuSO_4}=\dfrac{0,07}{0,1}=0,7\left(M\right)\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\\ Ba\left(OH\right)_2+2Zn+6H_2O\rightarrow Ba\left(Zn\left(OH\right)_4\right)_2+3H_2\)
\(\Rightarrow\)Dung dịch Y gồm \(Ba\left(OH\right)_2\left(dư\right)\) và \(Ba\left(Zn\left(OH\right)_4\right)_2\), khí Z là \(H_2\), chất rắn A gồm \(MgO,Fe_2O_3,Cu\)
\(H_2+MgO\rightarrow Mg+H_2O\\ Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)
\(\Rightarrow\)B gồm \(Mg,Fe,Cu\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\\ Mg+2HCl\rightarrow MgCl_2+H_2\)
\(\Rightarrow\)T gồm \(MgCl_2,FeCl_2,HCl\left(dư\right)\), C là Cu
\(FeCl_2+2HCl+4AgNO_3\rightarrow Fe\left(NO_3\right)_3+4AgCl+NO_2+H_2O\\ MgCl_2+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2AgCl\)
\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)
\(\Rightarrow\)Khí hóa nâu trong không khí là NO₂, E gồm AgCl
_{P/s: mình nghĩ chất rắn E chỉ có AgCl chứ còn chất nào kết tủa nữa đâu?:))}

\(Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{9,9161}{24,79}\approx0,4\left(mol\right)\\ b,m_{Fe}\approx0,4.56\approx22,4\left(g\right)\\ c,n_{HCl}\approx0,4.2\approx0,8\left(mol\right)\\ C_{MddHCl}\approx\dfrac{0,8}{0,25}\approx3,2\left(M\right)\)

\(a)Fe+2HCl\rightarrow FeCl_2+H_2\\ b)n_{H_2}=\dfrac{9,9161}{24,79}\approx0,4mol\\ n_{Fe}=n_{H_2}=0,4mol\\ m_{Fe}=0,4.56=22,4g\\ c)n_{HCl}=2n_{H_2}=2.0,4=0,8mol\\ 250ml=0,25l\\ C_{M_{HCl}}=\dfrac{0,8}{0,25}=3,2M\)

a) \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PTHH: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,15\cdot56=8,4\left(g\right)\)

c) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\)

\(V_{H_2SO_4}=50ml=0,05\left(l\right)\)

\(\Rightarrow C_{M,H_2SO_4}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,15}{0,05}=3M\)

\(a,Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,500ml=0,5l\\ n_{H_2SO_4}=0,5.1=0,5mol\\ n_{Zn}=\dfrac{13}{65}=0,2mol;n_{Fe}=\dfrac{14}{56}=0,25mol\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(0,2.....0,2...........0,2..........0,2\)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(0,25...0,25..........0,25......0,25\)
\(\Rightarrow n_{H_2SO_4.pứ}< n_{H_2SO_4}\left(0,2+0,25< 0,5\right)\\ \Rightarrow H_2SO_4.dư\\ V_{H_2}=\left(0,2+0,25\right)22,4=10,08l\\ c,m_{dd.lúc.đầu}=13+14+500.1,14=597g\\ m_{nước.thêm.vào}=600-597=3g\\ C_{\%H_2SO_4.dư}=\dfrac{\left(0,5-0,2-0,25\right).98}{600}\cdot100\%=0,82\%\\ C_{\%ZnSO_4}=\dfrac{0,2.161}{600}\cdot100\%=5,37\%\\ C_{\%FeSO_4}=\dfrac{0,25.152}{600}\cdot100\%=6,33\%\)