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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{Na}=\dfrac{4.6}{23}=0.2\left(mol\right)\)
\(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
\(0.2......................0.2.......0.1\)
\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)
Dung dịch X : NaOH
\(m_{dd_X}=4.6+200-0.1\cdot2=204.4\left(g\right)\)
\(C\%_{NaOH}=\dfrac{0.2\cdot40}{204.4}\cdot100\%=3.9\%\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
=> \(C\%_{dd.NaOH}=\dfrac{0,05.40}{20}.100\%=10\%\)
\(V_{dd.NaOH}=\dfrac{20}{1,2}=\dfrac{50}{3}\left(ml\right)=\dfrac{1}{60}\left(l\right)\)
\(C_{M\left(dd.NaOH\right)}=\dfrac{0,05}{\dfrac{1}{60}}=3M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
mH2SO4= \(\dfrac{300.7,35}{100}=22,05g\)
nH2SO4= \(\dfrac{22,05}{98}=0,225 mol\)
mHCl= \(\dfrac{200.7,3}{100}=14,6g\)
nHCl= \(\dfrac{14,6}{36,5}=0,4mol\)
H2SO4 + 2HCl → 2H2O + Cl2 ↑+ SO2 ↑
n trước pư 0,225 0,4
n pư 0,2 ← 0,4 → 0,4 → 0,2 → 0,2 mol
n sau pư dư 0,025 hết
a) mCl2= 0,2. 71= 14,2g
mSO2= 64. 0,2= 12,8g
mH2O= 18. 0,4=7,2g
mdd sau pư= 300 +200 -14,2 -12,8= 473g
C%dd H2O= \(\dfrac{7,2.100}{473}=1,52\)%
b) Mg + 2H2O → Mg(OH)2 + H2 ↑
x → 2x → x → x
Fe + 2H2O → Fe(OH)2 + H2↑
y → 2y → y → y
Gọi x,y lần lượt là số mol của Mg,Fe.
Ta có hệ phương trình:
24x + 56y = 8,7 x= \(\dfrac{5}{64}\)
⇒
2x + 2y = 0,4 y= \(\dfrac{39}{320}\)
VH2= 22,4. \((\dfrac{5}{64}+\dfrac{39}{320})\)= 4,48l
mhh MG(OH)2, Fe(OH)2= 8,7 +250 - 2.(\(\dfrac{5}{64}+\dfrac{39}{320}\)) = 2258,3g
mMg=24. \(\dfrac{5}{64}\)=1.875g
mFe= 8,7-1,875= 6,825g
![](https://rs.olm.vn/images/avt/0.png?1311)
nCO2=0,4(mol)
a) PTHH: 2 NaOH + CO2 -> Na2CO3 + H2O
0,8_________0,4________0,4(mol)
=> mNaOH=0,8.40=32(g)
=>C%ddNaOH=(32/200).100=16%
b) mddNa2CO3=mddNaOH+mCO2=200+0,4.44=217,6(g)
mNa2CO3=106.0,4=42,4(g)
=>C%ddNa2CO3=(42,4/217,6).100=19,485%
Chúc em học tốt!
nCO2=8,96/22,4=0,4mol
a/ CO2+2NaOH→Na2CO3+H2O
0,4 0,8 0,4 0,4
mNaOH=0,8.40=32g
C%ddNaOH=mct/mdd.100%=32/200.100%=16%
b/mCO2=0,4.44=17,6g
Theo định luật bảo toàn khối lượng:
mCO2+mNaOH=mNa2CO3
17,6g+200g=217,6g
mNa2CO3=0,4.106=42,4g
C%ddNa2CO3=mct/mdd.100%=42,4/217,6.100=19,4852g
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Chất rắn là MgO
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\)
0,8<-------------0,8<-------0,4
\(m=0,8.23+8=26,4\left(g\right)\)
\(b,m_{dd}=0,8.23+200-0,4.2=217,6\left(g\right)\\ \rightarrow C\%_{NaOH}=\dfrac{0,8.40}{217,6}.100\%=14,7\%\)
Tk:
a)2Al+ 6HCl→ 2AlCl3 +3H2↑
0,1________________0,15
Mg+ 2HCl→ MgCl2+ H2↑
0,2_______________0,2
MgO+ 2HCl→MgCl2+H2O
2Al+ 2NaOH+2H2O→ 2NaAlO2+ 3H2↑
0,1____________________________0,15
nHCl pư= 0,5.2.100110 =0,91 mol
nMgO=0,91−0,1.3−0,2.22= 0,105 mol
⇒ a= 0,1.27+0,2.24+ 0,105.40=11,7 g
b)
Dd B gồm:_______HCl dư ______AlCl3______MgCl2
_________________0,09________ 0,1_________0,305
NaOH+ HCl→ NaCl+ H2O
0,09 ___0,09
2NaOH+ MgCl2→ Mg(OH)2↓+ 2NaCl
0,61 ___0,305
3NaOH+ AlCl3→ 3NaCl+ Al(OH)3↓
0,3______ 0,1
VNaOH=0,09+0,61+0,32=0,5l
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
$2K + 2H_2O \to 2KOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)
Gọi $n_K = a(mol) ; n_{Na} = b(mol) \Rightarrow 39a + 23b = 8,5(1)$
Theo PTHH :
$n_{H_2} = 0,5a + 0,5b = \dfrac{3,36}{22,4} = 0,15(2)$
Từ (1)(2) suy ra a = 0,1 ; b = 0,2
$C_{M_{KOH}} = \dfrac{0,1}{0,2} = 0,5M$
$C_{M_{NaOH}} = \dfrac{0,2}{0,2} = 1M$
![](https://rs.olm.vn/images/avt/0.png?1311)
a) pthh: 2Na + 2H2O => 2NaOH + H2 (1)
Na2O + H2O => 2NaOH (2)
Sau khi xảy ra phản ứng sẽ sinh ra NaOH, vậy dd Y là dd NaOH
b) nH2 = 1.12/22.4 = 0.05 (mol)
theo pthh (1): nNa= nNaOH = 2nH2 = 2 x 0.05 = 0.1 (mol)
⇒ mNa phản ứng= mNa trong hh X = 0.1 x 23 = 2.3 g
mNa2O trong hh Y= 8.5 - 2.3 = 6.2g
c) mdd Y= mdd sau pứ= mhhX + mH2O - mH2 = 8.5 +200 -1.12=207.38g
nNa2O pứ = 6.2/62 = 0.1 (mol)
theo pthh (2): nNaOH = nNa2O pứ = 0.1 (mol)
nNaOH được sinh ra sau phản ứng = 0.1 +0.1 = 0.2 (mol)
mNaOH = 0.2 x 40 = 8g
Ta có C% =\(\dfrac{mct}{mdd}\times100\%\)
⇒ C%dd NaOH = \(\dfrac{8}{207.38}\times100\%\) = 3.86%
a) 2Na + 2H2O -> 2NaOH + H2
Na2O + H2O -> 2NaOH
ddY: NaOH
b) nH2 = 1.12/22.4 = 0.05mol
2Na + 2H2O -> 2NaOH + H2
(mol) 0.1 0.1 0.05
mNa = 0.1*23=2.3g
mNa2O = mhh - mNa = 8.5-2.3=6.2g
c)mH2 = 0.05*2=0.1g
mddY = mhh + mH2O - mH2
= 8.5 + 200 - 0.1= 208.4g
nNa2O = 6.2/62=0.1mol
Na2O + H2O -> 2NaOH
(mol) 0.1 0.2
nNaOH = 0.2+ 0.1 = 0.3mol
mNaOH = 0.3*40=12g
C% = 12/208.4*100%=5.76%
a)\(Ba+2H2O--->Ba\left(OH\right)2+H2\)
x----------------------------------------------x(mol)
\(2Na+2H2O-->2NaOH+H2\)
y-------------------------------------------0,5y(mol)
b) \(n_{H2}=\frac{1,344}{22,4}=0,06\left(mol\right)\)
Theo bài ra ta có hpt
\(\left\{{}\begin{matrix}137x+23y=6,4\\x+0,5y=0,06\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,04\end{matrix}\right.\)
Theo pthh1
n\(_{H2O}=2n_{Ba}=0,08\left(mol\right)\)
m\(_{H2O\left(1\right)}=0,08.18=1,44\left(g\right)\)
Theo pthh2
\(n_{H2O}=n_{Na}=0,04\left(mol\right)\)
\(m_{H2O}=0,04.18=0,72\left(g\right)\)
\(b=m_{H2O}=1,44+0,72=2,16\left(g\right)\)
\(n_{Ba\left(OH\right)2}=n_{Ba}=0,04\left(mol\right)\)
\(m_{Ba\left(OH\right)2}=0,04.171=6,84\left(g\right)\)
\(m_{ddBa\left(OH\right)2}=\frac{6,84.100}{3,42}=200\left(g\right)\)
do ở trong dd Y nên m dd Ba(OH)2 = m dd NaOH
n\(_{NaOH}=n_{Na}=0,04\left(mol\right)\)
m\(_{NaOH}=0,04.40=1,6\left(g\right)\)
\(C\%_{NaOH}=\frac{1,6}{200}.100\%=0,8\%\)