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\(n_{H_2SO_4}=0,25.2=0,5\left(mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)
\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)
x----------> 3x --------> x
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
y --------> y --------> y
Có hệ phương trình
\(\left\{{}\begin{matrix}102x+80y=26,2\\3x+y=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(\%_{m_{Al_2O_3}}=\dfrac{102.0,1.100}{26,2}=38,93\%\)
\(\%_{m_{CuO}}=\dfrac{80.0,2.100}{26,2}=61,07\%\)
\(CM_{Al_2\left(SO_4\right)_3}=\dfrac{x}{0,25}=\dfrac{0,1}{0,25}=0,4M\)
\(CM_{CuSO_4}=\dfrac{y}{0,25}=\dfrac{0,2}{0,25}=0,8M\)
a/ Gọi x,y lần lượt là số mol CuO và ZnO tham gia phản ứng
nHCl = 14,6/36,5 = 0,4 (mol)
PTHH : CuO + 2HCl -----> CuCl2 + H2O
(mol) x 2x x
ZnO + 2HCl -----> ZnCl2 + H2O
(mol) y 2y y
Ta có hệ pt : \(\begin{cases}80x+81y=16,08\\2x+2y=0,4\end{cases}\) \(\Leftrightarrow\begin{cases}x=0,12\\y=0,08\end{cases}\)
=> mCuO = 0,12.80 = 9,6 (g)
\(\Rightarrow\%CuO=\frac{9,6}{16,08}.100\approx59,7\%\)
=> %ZnO = 100% - 59,7% = 40,3%
b/ mCuCl2 = 0,12.135 = 16,2(g)
mZnCl2 = 0,08.136 = 10,88 (g)
PT: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(FeO+2HCl\rightarrow FeCl_2+H_2O\)
Gọi: \(\left\{{}\begin{matrix}n_{MgO}=x\left(mol\right)\\n_{FeO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 40x + 72y = 15,2 (1)
Ta có: nHCl = 0,3.2 = 0,6 (mol)
Theo PT: \(n_{HCl}=2n_{MgO}+2n_{FeO}=2x+2y=0,6\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{MgO}=0,2.40=8\left(g\right)\)
Bài 14 :
\(a) n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ CuO + 2HCl \to CuCl_2 + H_2O\\ n_{HCl} = 2n_{CuO} = 0,2(mol)\\ m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)\\ b) \text{Chất tan : } CuCl_2\\ n_{CuCl_2} = n_{CuO} = 0,1(mol)\\ m_{CuCl_2} = 0,1.135 = 13,5(gam)\)
Bài 15 :
\(a) n_{Fe_2O_3} =\dfrac{4,8}{160} = 0,03(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,09(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,09.98}{9,8\%} = 90(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,03(mol)\\ m_{Fe_2(SO_4)_3} = 0,03.400 = 12(gam)\)
\(CuO+2HCl \to CuCl_2+H_2O\\ Al_2O_3+6HCl \to 2AlCl_3+3H_2O\\ n_{CuO}=a(mol)\\ n_{Al_2O_3}=b(mol)\\ n_{HCl}=2a+6b=0,08(1)\\ m_{muối}=135a+267b=4,02(2)\\ (1)(2)\\ a=b=0,01(mol)\\ m_{dd}=4,02+0,01.(80+102)=5,84g\\ C\%_{CuO}=\frac{0,01.80}{5,84}.100\%=13,7\%\\ C\%_{Al_2O_3}=\frac{0,01.102}{5,84}.100\%=17,4\%\)
Đặt \(\hept{\begin{cases}x=n_{CuO}\\y=n_{Al_2O_3}\end{cases}}\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Theo phương trình \(\hept{\begin{cases}x=n_{CuCl_2}\\2y=n_{AlCl_3}\end{cases}}\)
\(\rightarrow135x+133,5.2y=4,02\left(1\right)\)
\(n_{HCl}=0,8.0,1=0,08mol\)
\(\rightarrow2x+6y=0,08\left(2\right)\)
Từ (1) và (2) => x = y = 0,01 mol
\(\rightarrow\%m_{CuO}=\frac{0,01.80}{0,01.80+0,01.102}.100\%=43,96\%\)
\(\rightarrow\%m_{Al_2O_3}=100-43,96\%=56,04\%\)
Đặt:\(\hept{\begin{cases}x=n_{CuO}\\y=n_{al_2O_3}\end{cases}}\)
PTHH:\(CuO+2HCL\rightarrow CuCl_2+H_2O\)
\(Al_2O_3+6HCL\rightarrow2Alcl_3+3H_2O\)
Theo phương trình: \(\hept{\begin{cases}x=n_{CuCl_2}\\2y=n_{AlCl_3}\end{cases}}\)
\(\rightarrow135x+133,5\cdot2y=4,02\left(1\right)\)
\(n_{HCL}=0,8\cdot0,1=0,08mol\)
\(\rightarrow2x+6y=0,08\left(2\right)\)
\(\Rightarrow x=y=0,01mol\)
\(\rightarrow\%m_{CuO}=\frac{0,01\cdot80}{0,01\cdot80+0,01\cdot102}\cdot100=43,96\%\)
\(\rightarrow\%m_{al_2O_3}=100\%-43,96\%=56,04\%\)
\(a.Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ b.n_{H_2SO_4}=0,22.1,25=0,275mol\\ n_{Fe_2O_3}=a;n_{CuO}=b\\ \Rightarrow\left\{{}\begin{matrix}3a+b=0,275\\160a+80b=16\end{matrix}\right.\\ \Rightarrow a=0,075;b=0,05\\ \%m_{Fe_2O_3}=\dfrac{0,075.160}{16}\cdot100=75\%\\ \%m_{CuO}=100-75=25\%\)
PT: \(Fe_2O_3+6HNO_3\rightarrow2Fe\left(NO_3\right)_3+3H_2O\)
\(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)
Giả sử: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\)
⇒ 160x + 80y = 24 (1)
Có: \(m_{HNO_3}=800.6,3\%=50,4\left(g\right)\Rightarrow n_{HNO_3}=\dfrac{50,4}{63}=0,8\left(mol\right)\)
Theo PT: \(n_{HNO_3}=6n_{Fe_2O_3}+2n_{CuO}=6x+2y\left(mol\right)\)
⇒ 6x + 2y = 0,8 (2)
Từ (1) và (2) ⇒ x = y = 0,1 (mol)
a, mFe2O3 = 0,1.160 = 16 (g)
mCuO = 8 (g)
b, Có: m dd sau pư = mX + m dd HNO3 = 24 + 800 = 824 (g)
Theo PT: \(\left\{{}\begin{matrix}n_{Fe\left(NO_3\right)_3}=2n_{Fe_2O_3}=0,2\left(mol\right)\\n_{Cu\left(NO_3\right)_3}=n_{CuO}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{Fe\left(NO_3\right)_2}=\dfrac{0,2.242}{824}.100\%\approx5,87\%\\C\%_{Cu\left(NO_3\right)_2}=\dfrac{0,1.188}{824}.100\%\approx2,28\%\end{matrix}\right.\)
Bạn tham khảo nhé!
\(n_{Fe_2O_3}=a\left(mol\right),n_{CuO}=b\left(mol\right)\)
\(m_X=160a+80b=24\left(g\right)\left(1\right)\)
\(n_{HNO_3}=\dfrac{800\cdot6.3\%}{63}=0.8\left(mol\right)\)
\(Fe_2O_3+6HNO_3\rightarrow2Fe\left(NO_3\right)_3+6H_2O\)
\(CuO+2HNO_3\rightarrow Cu\left(NO_3\right)_2+H_2O\)
\(n_{HNO_3}=6a+2b=0.8\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.1\)
\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right),m_{CuO}=8\left(g\right)\)
\(m_{dd}=24+800=824\left(g\right)\)
\(C\%_{Fe\left(NO_3\right)_3}=\dfrac{0.2\cdot242}{824}\cdot100\%=5.87\%\)
\(C\%_{Cu\left(NO_3\right)_2}=\dfrac{0.1\cdot188}{824}\cdot100\%=2.3\%\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(0.1........0.2..................0.1\)
\(n_{CuO}=\dfrac{13.6-0.1\cdot56}{80}=0.1\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
\(0.1.......0.2\)
\(C_{M_{HCl}}=\dfrac{0.2+0.2}{0.4}=1\left(M\right)\)
PT: \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Gọi: \(\left\{{}\begin{matrix}n_{Al_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 102x + 80y = 17,1 (1)
Ta có: \(m_{HCl}=300.7,3\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
Theo PT: \(n_{HCl}=6n_{Al_2O_3}+2n_{CuO}=6x+2y=0,6\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{CuO}=0,15.80=12\left(g\right)\)