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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/ CFeCl2=0,05/0,3=1/6 M
CFeCl3=0,1/0,3=1/3 M
CHCl 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
CNaOH=0,6/1,5=0,4M
Bài 1:
mdd NaOH = \(\dfrac{4}{10}.100=40\left(g\right)\)
mnước = mdd - mNaOH = 40 - 4 =36 (g)
Bài 2:
nH2 = \(\dfrac{1,12}{22,4}=0,05\) mol
Pt: 2K + 2H2O --> 2KOH + H2
0,1 mol<------------0,1 mol<-0,05 mol
.....K2O + H2O --> 2KOH
0,2 mol-----------> 0,4 mol
mK = 0,1 . 39 = 3,9 (g)
mK2O = mhh - mK = 22,7 - 3,9 = 18,8 (g)
=> nK2O = \(\dfrac{18,8}{94}=0,2\) mol
% mK = \(\dfrac{3,9}{22,7}.100\%=17,2\%\)
% mK2O = 100% - 17,2% = 82,8%
\(\sum n_{KOH}=0,1+0,4=0,5\left(mol\right)\)
CM KOH = \(\dfrac{0,5}{0,4}=1,25M\)
1.
\(\%C=\dfrac{12}{44}.100\simeq22,73\%\)
2.
\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ V=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(M\right)\)
\(C_{M\left(FeCl_2\right)}=\dfrac{n}{V}=\dfrac{0,05}{0,05}=1\left(M\right)\)
\(n_{BaCl_2}=\dfrac{41,6}{208}=0,2\left(mol\right)\)
\(n_{AgNO_3}=\dfrac{17}{170}=0,1\left(mol\right)\)
\(BaCl_2+2AgNO_3\rightarrow Ba\left(NO_3\right)_2+2AgCl\)
\(\dfrac{0,2}{1}>\dfrac{0,1}{2}\) ⇒ BaCl2 dư.
a, \(n_{AgCl}=n_{AgNO_3}=0,1\left(mol\right)\)
\(\Rightarrow m_{AgCl}=0,1.143,5=14,35\left(g\right)\)
b, \(n_{Ba\left(NO_3\right)_2}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\)
\(C_{M_{Ba\left(NO_3\right)_2}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)
\(n_{BaCl_2phan/ung}=\dfrac{1}{2}n_{AgNO_3}=0,05\left(mol\right)\)
\(\Rightarrow n_{BaCl_2dư}=0,15\left(mol\right)\rightarrow C_{M\left(BaCl_2\right)}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)
\(a.Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ n_{NaCl}=n_{HCl}=2.n_{CO_2}=2.\dfrac{448:1000}{22,4}=0,04\left(mol\right)\\ C_{MddHCl}=\dfrac{0,04}{0,02}=2\left(M\right)\\ b.m_{NaCl}=58,5.0,04=2,34\left(g\right)\\ c.m_{Na_2CO_3}=106.0,02=2,12\left(g\right)\\ \%m_{Na_2CO_3}=\dfrac{2,12}{5}.100=42,4\%\\ \%m_{NaCl}=100\%-42,4\%=57,6\%\)
Bài 16 :
\(n_{CO2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\)
Pt : \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O|\)
1 2 2 1 1
0,02 0,04 0,04 0,02
a) \(n_{HCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
20ml = 0,02l
\(C_{M_{HCl}}=\dfrac{0,04}{0,02}=2\left(M\right)\)
b) \(n_{NaCl}=\dfrac{0,02.2}{1}=0,04\left(mol\right)\)
⇒ \(m_{NaCl}=0,04.58,5=2,34\left(g\right)\)
c) \(n_{Na2CO3}=\dfrac{0,04.1}{2}=0,02\left(mol\right)\)
⇒ \(m_{Na2CO3}=0,02.106=2,12\left(g\right)\)
\(m_{NaCl}=5-2,12=2,88\left(g\right)\)
0/0Na2CO3 = \(\dfrac{2,12.100}{5}=42,4\)0/0
0/0NaCl = \(\dfrac{2,88.100}{5}=57,6\)0/0
Chúc bạn học tốt
CHÚC BẠN HỌC TỐT!!
Gọi số mol của FeO và Fe2O3 là a (mol)
Theo đề bài, ta có: \(m_{hh}=11,6\left(g\right)\)
\(\Rightarrow72a+160a=11,2\)
\(\Rightarrow a=0,05\left(mol\right)\)PTHH: \(FeO+2HCl\rightarrow FeCl_2+H_2O\)
pư............0,05...........0,1..............0,05..............0,05 (mol)
PTHH: \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)
pư.............0,05...........0,3.............0,1...............0,15 (mol)
b) \(\Rightarrow\left\{{}\begin{matrix}m_{FeO}=72.0,05=3,6\left(g\right)\\m_{Fe2O3}=11,6-3,6=8\left(g\right)\end{matrix}\right.\)
c) Ta có: \(n_{HCl\left(dư\right)}=0,2.3-\left(0,1+0,3\right)=0,2\left(mol\right)\)
\(\left\{{}\begin{matrix}C_{M\left(FeCl2\right)}=\dfrac{0,05}{0,2}=0,25\left(M\right)\\C_{MFeCl3}=\dfrac{0,1}{0,2}=0,5\left(M\right)\\C_{M\left(HCldư\right)}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)
Vậy..............
11,2 ở đâu ra vậy a cho e hỏi ???