Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{SO_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(n_{NaOH}=0.05\cdot1=0.05\left(mol\right)\)
\(\dfrac{n_{NaOH}}{n_{SO_2}}=\dfrac{0.05}{0.05}=1\)
\(X:NaHSO_3\)
- Xuất hiện kết tủa vàng nhạt
AlBr3 + 3AgNO3 => Al(NO3)3 +3AgBr
- Hỗn hợp tan dần, sủi bọt khí.
MgO + 2HCl => MgCl2 + H2O
CaCO3 + 2HCl => CaCl2 + CO2 + H2O
\(Đặt:n_{Fe}=a\left(mol\right),n_{FeO}=b\left(mol\right)\)
\(m_{hh}=56a+72b=9.2\left(g\right)\left(1\right)\)
\(n_{SO_2}=\dfrac{3.92}{22.4}=0.175\left(mol\right)\)
\(Fe^0\rightarrow Fe^{+3}+3e\)
\(Fe^{+2}\rightarrow Fe^{+3}+1e\)
\(S^{+6}+2e\rightarrow S^{+4}\)
\(\text{Bảo toàn electron: 3a + b = 0.35 (2)}\)
\(\left(1\right),\left(2\right):a=0.1,b=0.05\)
\(m_{Fe}=0.1\cdot56=5.6\left(g\right)\)
\(m_{FeO}=0.05\cdot72=3.6\left(g\right)\)
\(n_{H_2SO_4}=0.1\cdot3+0.05\cdot2=0.4\left(mol\right)\)
\(m_{H_2SO_4}=0.4\cdot98=39.2\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{39.2\cdot100}{98}=40\left(g\right)\)
\(\text{Mỗi phần,gọi :} n_{Al} = a ; n_{Fe} = b ; n_{Cu} = c\\ \Rightarrow 27a + 56b + 64c = \dfrac{35,8}{2} = 17,9(1)\\ \text{Phần 1 : Al,Fe không phản ứng với axit đặc nguội}\\ Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + H_2O\\ n_{Cu} = c = n_{SO_2} = \dfrac{3,36}{22,4} = 0,15(2)\\ \text{Phần 2 : Cu không phản ứng với axit loãng}\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ Fe + H_2SO_4 \to FeSO_4 + H_2\\ n_{H_2} = 1,5a + b = \dfrac{5,6}{22,4} = 0,25(3)\\ (1)(2)(3) \Rightarrow a = b = 0,1 ; c = 0,15\)
Suy ra :
\(m_{Al} = 0,1.2.27 = 5,4(gam)\\ m_{Fe} = 0,1.2.56 = 11,2(gam)\\ m_{Cu} = 0,15.64.2 = 19,2(gam)\)
Câu 2
\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)
\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)
\(a)n_{Mg} = a ; n_{Al} = b \Rightarrow 24a +27b = 5,1(1)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{H_2} = a + 1,5b = \dfrac{5,6}{22,4} = 0,25(2)\\ (1)(2) \Rightarrow a = 0,1 ; b = 0,1\\ \%m_{Mg} = \dfrac{0,1.24}{5,1}.100\% = 44,44\%\ ;\ \%m_{Al} = 100\% -44,44\% = 55,56\%\\ b) n_{MgCl_2} = n_{Mg} = 0,1 \Rightarrow m_{MgCl_2} = 0,1.95 = 9,5(gam)\\ n_{AlCl_3} = n_{Al} = 0,1 \Rightarrow m_{AlCl_3} = 0,1.133,5 = 13,35(gam)\\ c)n_{HCl} = 2n_{Mg} + 3n_{Al} = 0,5(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,5.36,5}{3,65\%} = 500(gam)\)
\(m_{dd\ sau\ pư} = 5,1 + 500 - 0,25.2 = 504,6(gam)\\ C\%_{MgCl_2} = \dfrac{9,5}{504,6}.100\% = 1,89\%\\ C\%_{AlCl_3} = \dfrac{13,35}{504,6}.100\% = 2,65\%\)
\(a) \\ 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ 3O_2 \xrightarrow{UV} 2O_3\\ O_3 + H_2O + 2KI \to 2KOH + O_2 + I_2\\ I_2 + 2K \xrightarrow{t^o} 2KI\\ 2KI + Br_2 \to 2KBr + I_2\\ 2KBr + Cl_2 \to 2KCl + Br_2\)
b)
\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ 2H_2O \xrightarrow{điện\ phân} 2H_2 + O_2\\ 3O_2 \xrightarrow{UV} 2O_3\\ 2Ag + O_3 \to Ag_2O + O_2\)
\(a) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ 3O_2 \xrightarrow{UV} 2O_3\\ 4O_3 + 9Fe \to 3Fe_3O_4\\ Fe_3O_4 + 8HCl \to 2FeCl_3 + FeCl_2 + 4H_2O\\ 2FeCl_2 + Cl_2 \to 2FeCl_3\\ b) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ 3O_2 \xrightarrow{UV} 2O_3\\ 2Ag + O_3 \to Ag_2O + O_2\\ Ag_2O + 2HNO_3 \to 2AgNO_3 + H_2O\\ AgNO_3 + HCl \to AgCl + HNO_3\)