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nKI = 83/166 = 0.5 (mol)
2KI + Cl2 => 2KCl + I2
0.5__0.25
2KMnO4 + 16HCl -to-> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0.1_______________________________0.25
Độ tinh khiết của KMnO4 ; 0.1*158/25 * 100% = 63.2%
\(n_{KI} = \dfrac{83}{166}=0,5(mol)\\ 2KI + Cl_2 \to 2KCl + I_2\\ n_{Cl_2} = \dfrac{n_{KI}}{2} = 0,25(mol)\\ 2KMnO_4 + 16HCl \to 2KCl + 2MnCl_2 + 5Cl_2 + 8H_2O\\ n_{KMnO_4\ pư} = \dfrac{2}{5}n_{Cl_2} = 0,1(mol)\\ \text{Độ tinh khiết} : \%m_{KMnO_4} = \dfrac{0,1.158}{25}.100\% = 63,2\%\)
\(a,2KMnO_4+16HCl_{đặc}\rightarrow\left(t^o\right)2KCl+2MnCl_2+5Cl_2+8H_2O\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ Ta.có:n_{FeCl_3}=\dfrac{39}{162,5}=0,24\left(mol\right)\\ n_{Fe}=n_{FeCl_3}=0,24\left(mol\right)\\ n_{Cl_2}=\dfrac{3}{2}.0,24=0,36\left(mol\right)\\ n_{K_2MnO_4}=\dfrac{2}{5}.0,36=0,144\left(mol\right)\\ n_{HCl}=\dfrac{16}{5}.0.36=1,152\left(mol\right)\\ \Rightarrow a=m_{KMnO_4}=0,144.158=22,752\left(g\right)\\ b=C_{MddHCl}=\dfrac{1,152}{0,1}=11,52\left(M\right)\\ x=m_{Fe}=0,24.56=13,44\left(g\right)\\ V=V_{Cl_2\left(đktc\right)}=0,36.22,4=8,064\left(l\right) \)
\(b,n_{KCl}=n_{MnCl_2}=\dfrac{2}{5}.0,36=0,144\left(mol\right)\\ KCl+AgNO_3\rightarrow AgCl\downarrow\left(trắng\right)+KNO_3\\ MnCl_2+2AgNO_3\rightarrow2AgCl\downarrow\left(trắng\right)+Mn\left(NO_3\right)_2\\ n_{AgNO_3}=n_{AgCl}=n_{KCl}+2.n_{MnCl_2}=0,144+2.0,144=0,432\left(mol\right)\\ \Rightarrow m_{AgCl\downarrow\left(trắng\right)}=143,5.0,432=61,992\left(g\right)\\ m_{AgNO_3}=0,432.170=73,44\left(g\right)\\ \Rightarrow m_{ddAgNO_3}=\dfrac{73,44.100}{5}=1468,8\left(g\right)\)
- PT: a, \(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(MnO_2+4HCl_đ\underrightarrow{t^o}MnCl_2+Cl_2+2H_2O\) (2)
- Ta có: \(n_{HCl\left(1\right)}=n_{HCl\left(2\right)}=0,2.2=0,4\left(mol\right)\)
Theo PT (1): \(n_{Cl_2}=\dfrac{5}{16}n_{HCl\left(1\right)}=0,125\left(mol\right)\Rightarrow V_1=0,125.22,4=2,8\left(l\right)\)
(2): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)
\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
\(Fe^{2+} \to Fe^{3+} + 1e\\ Mn^{+7} + 5e \to Mn^{2+}\\ \Rightarrow n_{Fe^{2+}} = 5n_{KMnO_4} = 0,18.5 =0,9(mol)\\ 2FeCl_3 + Fe \to 3FeCl_2\\ n_{FeCl_3} = \dfrac{2}{3}n_{FeCl_2} = 0,6(mol)\\ n_{Fe\ pư} = \dfrac{1}{3}n_{Fe} = 0,3(mol)\\ \Rightarrow m_{Fe\ trong\ A} = 2,8 + 0,3.56 = 19,6(gam)\\ 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ n_{Fe} = n_{FeCl_3} = 0,6(mol)\\\)
Phần trăm khối lượng Fe tham gia phản ứng là : \(\dfrac{0,6.56}{0,6.56 + 19,6}.100\% = 63,15\%\)
a ơi nhưng trong đề của cô e cho 4 đáp án không có đáp án 63,15%
- Theo bài ra \(\Rightarrow\left\{{}\begin{matrix}n_{KMnO_4}=0,1\\n_{KClO_3}=0,15\end{matrix}\right.\) ( mol )
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
.......0,1..........................................................0,25...........
\(KClO_3+6HCl\rightarrow KCl+3Cl_2+3H_2O\)
....0,15................................0,45....................
\(\Rightarrow n_{HCl}=0,7\left(mol\right)\)
\(6KOH+3Cl_2\rightarrow KClO_3+5KCl+3H_2O\)
Ta có : \(m=m_{KOH}+m_{Cl_2}=139,3\left(g\right)\)
Vậy ...
m giảm = mO2 = 0.8 (g)
BT e :
5nKMnO4 = 4nO2 + 2nCl2
=> nCl2 = (5*31.6/158 - 4*0.8/32)/2 = 0.45 (mol)
2NaOH + Cl2 => NaCl + NaClO + H2O
0.9______0.45
Vdd NaOH = 0.9/0.1 = 9 (l)
Em xem lại đáp án của đề nhé !!!