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1) \(6H_2O+6CO_2\rightarrow C_6H_{12}O_6+6O_2\)
2) \(2Fe+3Cl_2\rightarrow2FeCl_3\)
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)
a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)
Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)
\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)
\(a.Sắt+Clo\rightarrow Sắt\left(III\right)clorua\\ b.2Fe+3Cl_2-^{t^o}\rightarrow2FeCl_3\\ c.m_{Fe}+m_{Cl_2}=m_{FeCl_3}\\ \Rightarrow m_{FeCl_3}=5,6+10,65=16,25\left(g\right)\)
\(a,2Al+3Cl_2\rightarrow\left(t^o\right)2AlCl_3\\ b,Na_2CO_3+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+2NaOH\\ c,3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
\(a,2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\ 2:3:2\\ b,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ 3:2:1\\ c,K_2CO_3+H_2SO_4\to K_2SO_4+H_2O+CO_2\uparrow\\ 1:1:1:1:1\)
PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)
a, \(n_{Fe}=\dfrac{11}{56}\left(kmol\right)\)
Theo PT: \(n_{Fe_3O_4\left(LT\right)}=\dfrac{1}{3}n_{Fe}=\dfrac{11}{168}\left(kmol\right)\)
\(\Rightarrow m_{Fe_3O_4\left(LT\right)}=\dfrac{11}{168}.232=\dfrac{319}{21}\left(kg\right)\) > mFe3O4 (TT) = 200 (kg)
→ vô lý
Bạn xem lại đề phần a nhé.
b, \(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(kmol\right)\)
Theo PT: \(n_{Fe\left(LT\right)}=3n_{Fe_3O_4}=0,3\left(kmol\right)\)
\(\Rightarrow m_{Fe\left(LT\right)}=0,3.56=16,8\left(kg\right)\)
Mà: H = 85%
\(\Rightarrow m_{Fe\left(TT\right)}=\dfrac{16,8}{85\%}=\dfrac{336}{17}\left(kg\right)\)
\(n_{Fe}=\dfrac{6.72}{56}=0.12\left(mol\right)\)
\(n_{Cl_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(2Fe+3Cl_2\underrightarrow{^{t^0}}2FeCl_3\)
\(0.1........0.15....0.1\)
\(m_{Fe\left(dư\right)}=\left(0.12-0.1\right)\cdot56=1.12\left(g\right)\)
\(m_{FeCl_3}=0.1\cdot162.5=16.25\left(g\right)\)
\(a,\text{Sơ đồ p/ứ: }Fe+HCl\to FeCl_2+H_2\\ b,PTHH:Fe+2HCl\to FeCl_2+H_2\\ c,\text{Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}+56=150+8=158\\ \Rightarrow m_{HCl}=102(g)\)
