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\(n_{Cl_2}=\dfrac{2,1168}{22,4}=0,0945\left(mol\right)\)
=> nCl(muối) = 0,39 - 0,0945 = 0,201 (mol)
=> nAgCl = 0,201 (mol)
=> mAgCl = 0,201.143,5 = 28,8435 (g)
\(n_{KMnO_4}=\dfrac{15,8}{158}=0,1\left(mol\right)\)
PTHH: 2KMnO4 --to--> K2MnO4 + MnO2 + O2
a-------------->0,5a----->0,5a
=> 158(0,1-a) + 197.0,5a + 87.0,5a = 14,84
=> a = 0,06 (mol)
PTHH: 2KMnO4 + 16HCl --> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
0,04----------------------------------->0,1
K2MnO4 + 8HCl --> 2KCl + MnCl2 + 2Cl2 + 4H2O
0,03--------------------------------->0,06
MnO2 + 4HCl --> MnCl2 + Cl2 + 2H2O
0,03--------------------->0,03
=> \(n_{Cl_2}=0,1+0,06+0,03=0,19\left(mol\right)\)
=> \(V_{Cl_2}=0,19.22,4=4,256\left(l\right)\)
7)
nNaCl=x,nNaI=y
chất rắn sau nugn là NaCl n =1
x+y=1
58.5x + 150y=104.25
=>x=y=0.5
=> mNaCl=28.25 => % NaCl=28.06%
=> % NaI=71.49%
1)
nMnO2= 8,7/87 = 0,1 mol
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O
0,1 0,1 (mol)
V Cl2 = 0,1 x 22,4 = 2,24 lít
H%= 85% -> V Cl2 thu được = 2,24 x 85% = 1,904 lít
a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{Fe}=\dfrac{14}{56}=0,25\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=2n_{Fe}=0,5\left(mol\right)\\n_{H_2}=n_{Fe}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow V_{H_2}=0,25.24,79=6,1975\left(l\right)\)
\(a=C_{M_{HCl}}=\dfrac{0,5}{0,1}=5\left(M\right)\)
b, Theo PT: \(n_{FeCl_2}=n_{Fe}=0,25\left(mol\right)\)
Ta có: \(n_{AgNO_3}=0,4.1,3=0,52\left(mol\right)\)
PT: \(2AgNO_3+FeCl_2\rightarrow Fe\left(NO_3\right)_2+2AgCl_{\downarrow}\)
______0,5______0,25______0,25________0,5 (mol)
\(AgNO_3+Fe\left(NO_3\right)_2\rightarrow Fe\left(NO_3\right)_3+Ag_{\downarrow}\)
0,02______0,02________0,02________0,02 (mol)
⇒ m = mAgCl + mAg = 0,5.143,5 + 0,02.108 = 73,91 (g)
- Dd sau pư gồm: Fe(NO3)3: 0,02 (mol) và Fe(NO3)2: 0,25 - 0,02 = 0,23 (mol)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{Fe\left(NO_3\right)_3}}=\dfrac{0,02}{0,1+0,4}=0,04\left(M\right)\\C_{M_{Fe\left(NO_3\right)_2}}=\dfrac{0,23}{0,1+0,4}=0,46\left(M\right)\end{matrix}\right.\)
\(Fe+2HCl->FeCl_2+H_2\\ a.V=\dfrac{14}{56}\cdot22,4=5,6\left(L\right)\\ a=\dfrac{\dfrac{14}{56}\cdot2}{0,1}=5\left(M\right)\\ b.n_{AgNO_3}=0,4\cdot1,3=0,52mol\\ FeCl_2+AgNO_3->Fe\left(NO_3\right)_2+AgCl\\ Fe\left(NO_3\right)_2+AgNO_3->Ag+Fe\left(NO_3\right)_3\\ m=0,25\cdot143,5+0,25\cdot108=62,875\left(g\right)\\ C_{M\left(AgNO_3\right)}=\dfrac{0,02}{0,5}=0,04M\\ C_{M\left(Fe\left(NO_3\right)_3\right)}=\dfrac{0,25}{0,5}=0,5M\)
40. Đặt \(n_{Cl_2\left(thu\text{ được}\right)}=x\left(mol\right)\)
\(2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\)
___________x______x________x
\(\Rightarrow m_{muối}=133x=47,88\\ \Rightarrow x=0,36\left(mol\right)\)
\(n_{K_2Cr_2O_7}=0,15\left(mol\right)\)
QT nhận e: Cr2+6 + 6e ----> 2Cr+3
________0,15_____0,9
QT nhường e: 2Cl- ----> Cl2 + 2e
_____________________0,45___0,9
\(\Rightarrow HSPU=\frac{n_{Cl_2\left(thu\text{ được}\right)}}{n_{Cl_2\left(lý\text{ thuyết}\right)}}=80\%\)
42.
\(n_{KMnO_4}=0,2\left(mol\right);n_{NaCl}=0,375\left(mol\right)\)
\(2KMnO_4+16HCl\rightarrow2KCl+2MnCl_2+5Cl_2+8H_2O\)
0,2____________________________________0,5
\(2NaOH+Cl_2\rightarrow NaCl+NaClO\)
_________0,375___0,375
\(\Rightarrow H=\frac{0,375}{0,5}=75\%\)
\(a\text{) }n_{KMnO_4}=0,2\left(mol\right)\\ BT\text{ }e\Rightarrow5n_{KMnO_4}=2n_{Cl_2}\\ \Rightarrow n_{Cl_2}=0,5\left(mol\right)\Rightarrow V=11,2\left(l\right)\)
b) 2NaOH + Cl2 ---> NaCl + NaClO + H2O
______1________0,5_______0,5______0,5
\(\Rightarrow C_{M\left(NaOH\right)}=0,5\left(M\right);C_{M\left(NaCl\right)}=1,25\left(M\right);C_{M\left(NaClO\right)}=1,25\left(M\right)\)