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\(Đặt:n_{MnO_2}=a\left(mol\right),n_{KMnO_4}=b\left(mol\right)\)
\(m_{hh}=87a+158b=37.96\left(g\right)\left(1\right)\)
\(n_{Cl_2}=\dfrac{10.08}{22.4}=0.45\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
\(MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O\)
\(n_{Cl_2}=a+2.5b=0.45\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.4,b=0.02\)
\(\%MnO_2=\dfrac{0.4\cdot87}{37.96}\cdot100\%=91.68\%\\\%KMnO_4=100-91.68=8.32\% \)
\(m_M=m_{KCl}+m_{MnCl_2}=0.02\cdot74.5+\left(0.4+0.02\right)\cdot126=54.41g\)
nH2 = 0,672/22,4 = 0,03 (mol)
PTHH: Fe + 2HCl -> FeCl2 + H2
nFe = 0,03 (mol)
mFe = 0,03 . 56 = 1,68 (g)
mCu = 2,96 - 1,68 = 1,28 (g)
nCu = 1,28/64 = 0,02 (mol)
PTHH:
2Fe + 3Cl2 -> (t°) 2FeCl3
0,03 ---> 0,045
Cu + Cl2 -> (t°) CuCl2
0,02 ---> 0,02
nCl2 = 0,045 + 0,02 = 0,065 (mol)
VCl2 = 0,065 . 22,4 = 1,456 (l)
a: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6 0,2 0,3
\(V=0.3\cdot22.4=6.72\left(lít\right)\)
b: \(C_{M\left(HCL\right)}=\dfrac{0.6}{0.2}=3\left(M\right)\)
- 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): \(n_{Cl_2\left(2\right)}=\dfrac{1}{4}n_{HCl\left(2\right)}=0,1\left(mol\right)\Rightarrow V_2=0,1.22,4=2,24\left(l\right)\)
\(n_{KMnO_4}=\frac{15,8}{158}=0,1\left(mol\right)\)
PTHH : \(2KMnO_4+16HCl-->2KCl+2MnCl_2+5Cl_2+8H_2O\) (1)
\(Cl_2+H_2-as->2HCl\) (2)
Có : \(m_{ddHCl}=100\cdot1,05=105\left(g\right)\)
=> \(m_{HCl}=105-97,7=7,3\left(g\right)\)
=> \(n_{HCl}=\frac{7,3}{36,5}=0,2\left(mol\right)\)
BT Clo : \(n_{Cl_2}=\frac{1}{2}n_{HCl}=0,1\left(mol\right)\)
Mà theo lí thuyết : \(n_{Cl_2}=\frac{5}{2}n_{KMnO_4}=0,25\left(mol\right)\)
=> \(H\%=\frac{0,1}{0,25}\cdot100\%=40\%\)
Vì spu nổ thu được hh hai chất khí => \(\hept{\begin{cases}H_2\\HCl\end{cases}}\) (Vì H2 dư)
=> \(n_{hh}=\frac{13,44}{22,4}=0,6\left(mol\right)\)
=> \(n_{H_2\left(spu\right)}=n_{hh}-n_{HCl\left(spu\right)}=0,6-0,2=0,4\left(mol\right)\)
BT Hidro : \(\Sigma_{n_{H2\left(trong.binh\right)}}=n_{H_2\left(spu\right)}+\frac{1}{2}n_{HCl}=0,4+0,1=0,5\left(mol\right)\)
đọc thiếu đề câu a wtf
\(C_{M\left(HCl\right)}=\frac{0,2}{0,1}=2\left(M\right)\)
\(MnO_2+4HCl\underrightarrow{^{to}}MnCl_2+Cl_2+2H_2O\)
\(n_{MnO2}=\frac{8,7}{87}=0,1\left(mol\right)\)
\(\Rightarrow n_{Cl2}=n_{MnO2}=0,1\left(mol\right)\)
\(\Rightarrow V_{Cl2}=0,1.22,4=2,24\left(l\right)\)
\(n_{HCl}=2n_{MnO2}=4.0,1=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(\Rightarrow m_{dd_{HCl}}=\frac{14,6.100}{20}=73\left(g\right)\)
\(Cl_2+2NaOH\rightarrow NaCl+NaClO+H_2O\)
0,1_____0,2________0,1______0,1_____
\(n_{NaOH}=0,2.3=0,6\left(mol\right)\)
\(n_{Cl2}=0,1\left(mol\right)\)
\(\Rightarrow\) NaOH dư \(\Rightarrow\) nNaOH dư= 0,6-0,2=0,4 (mol)
Dung dịch sau phản ứng gồm: NaCl, NaClO và NaOH dư.
\(CM_{NaCl}=CM_{NaClO}=\frac{0,1}{0,2}=0,5M\)
\(CM_{NaOH_{Dư}}=\frac{0,4}{0,2}=2M\)