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a)
\(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)
PTHH: 2Na + 2H2O --> 2NaOH + H2
0,03<------------0,03<----0,015
=> \(\%m_{Na}=\dfrac{0,03.23}{1,31}.100\%=52,67\%\)
=> \(\%m_{Na_2O}=100\%-52,67\%=47,33\%\)
b)
\(n_{Na_2O}=\dfrac{1,31.47,33\%}{62}=0,01\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
0,01----------->0,02
=> nNaOH = 0,03 + 0,02 = 0,05 (mol)
mdd sau pư = 1,31 + 18,72 - 0,015.2 = 20 (g)
=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)
\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)
2a.
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
2Na + 2H2O \(\rightarrow\) 2NaOH + H2 (1)
K2O + H2O \(\rightarrow\) 2KOH (2)
Có : mdd B = mhh A + mH2O - mH2 = 19,85 + 180,4 - mH2 = 200
\(\Rightarrow\) mH2 = 0,25(g)
\(\Rightarrow\) nH2 = 0,25/2 = 0,125(mol)
Theo PT(1) \(\Rightarrow\)nNa = nNaOH = 2.nH2 = 2. 0,125 = 0,25(mol)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,25.23=5,75\left(g\right)\\m_{NaOH}=0,25.40=10\left(g\right)\end{matrix}\right.\)
\(\Rightarrow\) mK2O = 19,85 - 5,75= 14,1(g)
\(\Rightarrow\) nK2O = 14,1/94 = 0,15(mol)
Theo PT(2) \(\Rightarrow\) nKOH = 2 . nK2O = 2. 0,15 = 0,3(mol)
\(\Rightarrow\) mKOH = 0,3 . 56 = 16,8(g)
* C%KOH / ddB = 16,8/200 . 100% = 8,4%
C%NaOH / dd B = 10/200 . 100% = 5%
* m(KOH+ NaOH) = 16 ,8 + 10 =\ 26,8(g)
\(\Rightarrow\)mH2O / dd B = 200 - 26,8 = 173,2 (g)
\(\Rightarrow\) VH2O / dd B = m : D = 173,2 : 1 = 173,2 (ml) =0,1732(l)
mà Vdd B = VH2O / ddB
=> Vdd B =\ 0,1732(l)
Do đó :
CM của NaOH / dd B = 0,25/0,1732=1,44(M)
CM của KOH / dd B = 0,3/0,1732 = 1,73 (M)
\(m_{H_2} = 8,5 + 50 - 58,4 = 0,1(gam)\\ n_{H_2} = \dfrac{0,1}{2} = 0,05(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ Na_2O + H_2O \to 2NaOH\\ n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)\\ \Rightarrow n_{Na_2O} = \dfrac{8,5-0,1.23}{62}=0,1(mol)\\ n_{NaOH} = 2n_{Na_2O} + n_{Na} = 0,3(mol)\\ C\%_{NaOH} = \dfrac{0,3.40}{58,4}.100\% = 20,55\%\)
Hai bạn làm sai một số chỗ, mình sẽ làm lại
Bài 1:
\(Na_2O\left(0,1\right)+H_2O--->2NaOH\left(0,2\right)\)
\(n_{Na_2O}=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,2.40=8\left(g\right)\)
\(m_{ddsau}=6,2+73,8=80\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{8}{80}.100=10\%\)
Bài 2:
\(n_{Na_2O}=0,1\left(mol\right)\)
\(m_{NaOH}\left(bđ\right)=60\left(g\right)\)
\(\Rightarrow m_{H_2O}=133,8-60=73,8\left(g\right)\)\(\Rightarrow n_{H_2O}=4,1\left(mol\right)\)
\(Na_2O\left(0,1\right)+H_2O\left(0,1\right)--->2NaOH\left(0,2\right)\)
So sánh: \(\dfrac{n_{Na_2O}}{1}=0,1< \dfrac{n_{H_2O}}{1}=4,1\)
=> Chọn số mol của Na2O để tính
Theo PTHH: nNaOH (tạo thành) = 0,2 (mol)
=> mNaOH (tạo thành) = 8 (g)
\(\Rightarrow\sum m_{NaOH}\left(sau\right)=60+8=68\left(g\right)\)
\(m_{ddsau}=6,2+133,8=140\left(g\right)\)
\(\Rightarrow C\%_{NaOH}\left(sau\right)=\dfrac{68}{140}.100=48,57\%\)
Bài 3:
\(m_{NaOH}\left(bđ\right)=12\left(g\right)\)
\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)
\(Na_2O\left(\dfrac{a}{62}\right)+H_2O--->2NaOH\left(\dfrac{a}{31}\right)\)
\(m_{NaOH}\left(tao.thanh\right)=\dfrac{a}{31}.40=\dfrac{40a}{31}\left(g\right)\)
\(\Rightarrow\sum m_{NAoh}\left(sau\right)=12+\dfrac{40a}{31}\left(g\right)\)
\(m_{ddsau}=\left(a+120\right)\left(g\right)\)
Ta có: \(20=\dfrac{12+\dfrac{40a}{31}}{a+120}.100\)
\(\Rightarrow a=11\left(g\right)\)
BT 1:
mdd = mct + mdm = 6,2 + 73,8 = 80 (g)
C%A = \(\dfrac{m_{ct}}{m_{dd}}.100=\dfrac{6,2}{80}.100=7,75\%\)
\(n_{Na}=x\left(mol\right)\)
\(n_{K_2O}=y\left(mol\right)\)
\(m_{hhA}=23x+94y=18,7\left(I\right)\)
PTHH:
2Na + 2H2O \(\rightarrow\) 2NaOH + H2\(\uparrow\) (1)
(mol) x...........................x..............0,5x
K2O + H2O \(\rightarrow\) 2KOH (2)
(mol) y..........................y
\(m_{hhX}=m_{H_2O}+m_{hhA}-m_{H_2\uparrow}\)
\(200=181,5+18,7-m_{H_2\uparrow}\)
\(m_{H_2\uparrow}=0,2\left(g\right)\)
\(n_{H_2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)
\(\left(1\right)\rightarrow n_{H_2}=0,5.x=0,1\)
\(\rightarrow x=0,2\left(mol\right)\)
\(\left(I\right)\rightarrow y=0,15\left(mol\right)\)
200(g) ddX có 2 chất tan: NaOH, KOH
\(\left(1\right)\rightarrow n_{NaOH}=x=0,2\left(mol\right)\)
\(\left(2\right)\rightarrow n_{KOH}=2y=0,3\left(mol\right)\)
\(C\%_{NaOH/_{ddX}}=\dfrac{40.0,2}{200}.100=4\%\)
\(C\%_{KOH/_{ddX}}=\dfrac{56.0,3}{200}.100=8,4\%.\)