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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)
=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)
\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)
a, \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,2------------------>0,4
\(\Rightarrow C\%_{NaOH}=\dfrac{0,4.40}{12,4+50}.100\%=25,64\%\)
b, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
0,2------------------->0,2------->0,1
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40+16}{100+16+4,6-0,1.2}.100\%==20\%\)
c, \(n_{Na}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
\(n_{HCl}=\dfrac{100.7,3\%}{36,5}=0,2\left(mol\right)\)
PTHH:
2Na + 2HCl ---> 2NaCl + H2
0,2<-----0,2-----------0,2--->0,1
2Na + 2H2O ---> 2NaOH + H2
0,2------------------>0,2----->0,1
\(\Rightarrow m_{dd}=9,2+100-\left(0,1+0,1\right).2=108,8\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,2.58,5}{108,8}.100\%=10,75\%\\C\%_{NaOH}=\dfrac{0,2.40}{108,8}.100\%=7,35\%\end{matrix}\right.\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,1\left(mol\right)\)
m dd sau pư = 3,1 + 50 = 53,1 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{53,1}.100\%\approx7,53\%\)
b, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\end{matrix}\right.\)
Ta có: m dd sau pư = 4,6 + 95,6 - 0,1.2 = 100 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{100}.100\%=8\%\)
\(a,C\%_{KOH}=\dfrac{28}{140}.100\%=20\%\\ b,C\%_{KOH}=\dfrac{80}{80+320}.100\%=20\%\)
Sửa đề: 9,2 gam Na
\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)
PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)
0,4------------------>0,8
\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)
\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)
PTHH: \(K_2O+H_2O\rightarrow2KOH\)
0,4----------------->0,8
\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)
a, \(n_K=\dfrac{3,9}{39}=0,1\left(mol\right)\)
PTHH: 2K + 2H2O ---> 2KOH + H2
0,1---------------->0,1----->0,05
\(m_{ct}=m_{KOH}=0,1.56=5,6\left(g\right)\\ m_{dd}=m_K+m_{H_2O}-m_{H_2}=96,2+3,9-0,05.2=100\left(g\right)\)
\(C\%_{KOH}=\dfrac{5,6}{100}.100\%=5,6\%\\ b,m_{dd}=100+50=150\left(g\right)\\ C\%_{KOH}=\dfrac{5,6}{150}.100\%=3,37\%\)
c, Gọi \(m_{H_2O}=a\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\dfrac{5,6}{100+a}.100\%=2,8\%\\ \Leftrightarrow a=100\left(g\right)\)
d, Gọi \(m_{KOH}=a\left(g\right)\)
\(\Rightarrow C\%_{KOH}=\dfrac{5,6+a}{100+a}.100\%=22,4\%\\ \Leftrightarrow a=21,65\left(g\right)\)
\(n_{Na}=x\left(mol\right)\)
\(n_{K_2O}=y\left(mol\right)\)
\(m_{hhA}=23x+94y=18,7\left(I\right)\)
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
\(\left(1\right)\rightarrow n_{NaOH}=x=0,2\left(mol\right)\)
\(\left(2\right)\rightarrow n_{KOH}=2y=0,3\left(mol\right)\)
\(C\%_{NaOH/_{ddX}}=\dfrac{40.0,2}{200}.100=4\%\)
\(C\%_{KOH/_{ddX}}=\dfrac{56.0,3}{200}.100=8,4\%.\)
nNa = 6.9 : 23 = 0.3 mol
4Na + O2 ->2 Na2O
mol : 0.3 -> 0.15
Na2O + H2O -> 2NaOH
mol : 0.15 -> 0.3
mdd = 0.15 x 62 + 140.7 = 150g
C% NaOH = 0.3x40: 150 x 100% = 8%
\(m_{H_2} = 8,5 + 50 - 58,4 = 0,1(gam)\\ n_{H_2} = \dfrac{0,1}{2} = 0,05(mol)\\ 2Na + 2H_2O \to 2NaOH + H_2\\ Na_2O + H_2O \to 2NaOH\\ n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)\\ \Rightarrow n_{Na_2O} = \dfrac{8,5-0,1.23}{62}=0,1(mol)\\ n_{NaOH} = 2n_{Na_2O} + n_{Na} = 0,3(mol)\\ C\%_{NaOH} = \dfrac{0,3.40}{58,4}.100\% = 20,55\%\)