Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\left\{{}\begin{matrix}n_{FeO}=nFe2O3\\72.nFeO=160.nFe2O3\end{matrix}\right.=17,44\)
\(\Rightarrow n_{FeO}=n_{Fe2O3}=0,075\left(mol\right)\)
PTHH: FeO+2HCl-->FeCl2+H2O
...........0,075.....0,15.....0,075.............(mol)
Fe2O3+6HCl-->2FeCl3+3H2O
0,075......0,45................................(mol)
a, \(CM_{FeCl2}=\frac{0,075}{0,45}=0,17\left(M\right)\)
\(CM_{FeCl3}=\frac{0,15}{0,45}=0,33\left(M\right)\)
b, FeCl2+2NaOH-->Fe(OH)2+2NaCl
FeCl3+3NaOH-->Fe(OH)3+3NaCl
\(\Rightarrow n_{NaOH}=0,15+0,45=0,6\left(mol\right)\)
\(\Rightarrow V_{NaOH}=\frac{0,6}{2,5}=0,24l\)
Bài 1: \(n_{H_2SO_4}=\frac{9}{49}\left(mol\right)\)
H2SO4 + 2KOH -> K2SO4 + 2H2O
=> nKOH= 2nH2SO4 = \(\frac{18}{49}\left(mol\right)\)
=> Vdd KOH = \(\frac{18}{49}:\frac{2}{1000}=\frac{9000}{49}\left(ml\right)\)
b) nK2SO4 = nH2SO4 = \(\frac{9}{49}\left(mol\right)\)
=> mK2SO4= \(\frac{9}{49}\cdot174=\frac{1566}{49}\left(g\right)\)
mdd KOH = \(\frac{9000}{49}\cdot1,12=\frac{1440}{7}\left(g\right)\)
c) \(\%m_{K_2SO_4}=\frac{1566}{49}:\left(200+\frac{1440}{7}\right)\cdot100\%\approx7,87\%\)
bài 2: nNa2CO3 = 0,05 (mol)
PTHH:
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
=> nHCl = n NaCl = 2nNa2CO3 = 0,1 (mol)
=> mNaCl= 0,1 . 58,5 = 5,85 (g)
b) nCO2 = nNa2CO3 = 0,05 (mol)
=> mCO2 = 0,05 . 44 = 2,2 (g)
mdd HCl = 0,1 . 36,5 :20% = 18,25 (g)
=> %mNaCl = \(\frac{5,85}{53+18,25-2,2}\approx8,47\%\)
1)
a dd KOH
MgCl2 + 2KOH --------> Mg(OH)2 + 2KCl
Cu(NO3)2 + 2KOH ------> Cu(OH)2 + 2KNO3
b) AgNO3
2AgNO3 + MgCl2 -------> 2AgCl + Mg(NO3)2
nNa2O=15,5/62=0,25mol
pt : Na2O + H2O ---------> 2NaOH
npứ: 0,25---------------------->0,5
CM(NaOH)=0,5/0,5=1M
pt : 2NaOH + H2SO4 ------> Na2SO4 + 2H2O
npứ:0,5---------->0,25
mH2SO4 = 0,25.98=24,5g
mddH2SO4 =\(\dfrac{24,5.100}{20}=122,5\)
Vdd H2SO4=122,5/1,14\(\approx107,46ml\)
\(n_{FeO}=\dfrac{7,2}{72}=0,1mol\\ n_{H_2SO_4}=0,4.1,5=0,6mol\\ FeO+H_2SO_4\rightarrow FeSO_4+H_2O\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,6}{1}\Rightarrow H_2SO_4.dư\\ n_{FeO}=n_{FeSO_4}=n_{H_2SO_4,pư}=0,1mol\\ C_{M_{FeSO_4}}=\dfrac{0,1}{0,4}=0,25M\\ C_{M_{H_2SO_4}}=\dfrac{0,6-0,1}{0,4}=1,25M\)
nHCl=0,6 mol
FeO+2HCl-->FeCl2+ H2O
x mol x mol
Fe2O3+6HCl-->2FeCl3+3H2O
x mol 2x mol
72x+160x=11,6
=>x=0,05 mol
a) CMFeCl2=0,05/0,3=1/6 M
CM FeCl3=0,1/0,3=1/3 M
CM HCl du=(0,6-0,4)/0,3=2/3 M
b/
NaOH+ HCl-->NaCl+H2O
0,2 0,2
2NaOH+FeCl2-->2NaCl+Fe(OH)2
0,1 0,05
3NaOH+FeCl3-->3NaCl+Fe(OH)3
0,3 0,1
nNaOH=0,6
VNaOH=0,6/1,5=0,4l=400ml
Gọi nNa2CO3 = x (mol)
Na2CO3 + 2HCl \(\rightarrow\) 2NaCl + H2O + CO2
x \(\rightarrow\) 2x \(\rightarrow\) 2 x (mol)
C%(NaCl) = \(\frac{2.58,5x}{200+120}\) . 100% = 20%
=> x =0,547 (mol)
mNa2CO3 = 0,547 . 106 = 57,982 (g)
mHCl = 2 . 0,547 . 36,5 =39,931 (g)
C%(Na2CO3) =\(\frac{57,892}{200}\) . 100% = 28,946%
C%(HCl) = \(\frac{39,931}{120}\) . 100% = 33,28%
\(n_{HCl}=0,25.0,2=0,05\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a) Theo PT: \(\Rightarrow m_{Fe}=\frac{0,05}{2}.56=1,4\left(g\right)\)
b) Theo PT: \(\Rightarrow V_{H_2}=\frac{0,05}{2}.22,4=0,56\left(l\right)\)
c)Theo PT: \(n_{FeCl_2}=\frac{1}{2}n_{H_2}=0,025\left(mol\right)\)
\(\Rightarrow C_{M-FeCl_2}=\frac{0,025}{0,25}=0,1M\)