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Đáp án: A
Vì dung dịch rượu gồm rượu etylic và nước nên ta gọi:
n H 2 O = x m o l và n C 2 H 5 O H = y m o l
PTHH:
2 N a + 2 H 2 O → 2 N a O H + H 2 ↑ ( 1 )
x mol → 0,5.x mol
2 N a + 2 C 2 H 5 O H → 2 C 2 H 5 O N a + H 2 ↑
y mol → 0,5.y mol
Ta có hệ phương trình:
18 x + 46 y = 10 , 1 0 , 5 x + 0 , 5 y = 0 , 125 ⇒ x = 0 , 05 y = 0 , 2
V C 2 H 5 O H nguyên chất = m D = 0 , 2 . 46 0 , 8 = 11 , 5 m l
V H 2 O = m D = 10 , 1 - 9 , 2 1 = 0 , 9 m l
=> V d d r ư ợ u = V H 2 O + V C 2 H 5 O H = 0,9 + 11,5 = 12,4 ml
=> Độ rượu D 0 = V C 2 H 5 O H V d d r u o u . 100 = 11 , 5 12 , 4 . 100 = 92 , 74 0
a)
\(V_{C_2H_5OH}=\dfrac{96.20}{100}=19,2\left(ml\right)\)
=> \(m_{C_2H_5OH}=19,2.0,8=15,36\left(g\right)\)
b) \(n_{C_2H_5OH}=\dfrac{15,36}{46}=\dfrac{192}{575}\left(mol\right)\)
\(V_{H_2O}=20-19,2=0,8\left(ml\right)\)
=> \(m_{H_2O}=0,8.1=0,8\left(g\right)\)
=> \(n_{H_2O}=\dfrac{0,8}{18}=\dfrac{2}{45}\left(mol\right)\)
PTHH: 2C2H5OH + 2Na --> 2C2H5ONa + H2
\(\dfrac{192}{575}\)------------------------->\(\dfrac{96}{575}\)
2H2O + 2Na --> 2NaOH + H2
\(\dfrac{2}{45}\)----------------------->\(\dfrac{1}{45}\)
=> \(V_{H_2}=22,4.\left(\dfrac{96}{575}+\dfrac{1}{45}\right)=4,238\left(l\right)\)
C2H5OH + Na -- > C2H5OHNa + 1/2 H2
Na+H2O --- > NaOH + 1/2H2
Vr = 20x96/100 = 19,2ml = 0.0192 (l)
mC2H5OH = D.V = 19,2 x 0.8 = 15.36 (g)
nC2H5OH = m/M = 15.36 / 46 = 0.43 (mol)
=> nH2 = 0.215 (mol)
VH2O = 1 ml => mH2O = 1 (g)
=> nH2O = m/M = 1/18 = 0.056 (mol)
=> nH2 = 0.028 (mol)
nH2 = 0.215 + 0.028 = 0.243 (mol)
=> VH2 = 22.4 x 0,243 = 5,4432 (l)
a. \(V_{dd}=92+108=200\left(ml\right)\)
\(Đ_{rượu}=\dfrac{92}{200}.100=46^o\)
b.\(V_{dd\left(15^o\right)}=\dfrac{92.100}{11,5}=800\left(ml\right)\)
\(V_{H_2O\left(thêm\right)}=800-200=600\left(ml\right)\)
c.\(V_{rượu\left(23^o\right)}=\dfrac{100.23}{100}=23\left(ml\right)\)
\(V_{rượu\left(sau\right)}=23+92=115\left(ml\right)\)
\(V_{dd\left(sau\right)}=800+100=900\left(ml\right)\)
\(Đ_{rượu}=\dfrac{115}{900}.100=12,78^o\)
TN1: \(n_{H_2}=\dfrac{0,112}{22,4}=0,005\left(mol\right)\)
PTHH: 2C2H5OH + 2Na --> 2C2H5ONa + H2
0,01<-----------------------0,005
=> m = 0,01.46 = 0,46 (g)
TN2:
PTHH: C2H5OH + 3O2 --to--> 2CO2 + 3H2O
0,01----------------->0,02
=> nCO2 = 0,02 (mol)
nCa(OH)2 = 0,025.0,8 = 0,02 (mol)
Xét \(\dfrac{n_{OH^-}}{n_{CO_2}}=\dfrac{0,04}{0,02}=2\) => Pư vừa đủ, tạo ra muối CaCO3
CaCO3 kết tủa thì sao thu được dd được nhỉ :)
PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O_{ }\)
\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3\downarrow+H_2O\)
a) Ta có: \(n_{CaCO_3}=\dfrac{160}{100}=1,6\left(mol\right)=n_{CO_2}\) \(\Rightarrow n_{O_2}=2,4\left(mol\right)\)
\(\Rightarrow V_{kk}=\dfrac{2,4\cdot22,4}{20\%}=268,8\left(l\right)\)
b) Theo PTHH: \(n_{C_2H_5OH}=\dfrac{1}{2}n_{CO_2}=0,8\left(mol\right)\)
\(\Rightarrow a=C\%_{C_2H_5OH}=\dfrac{0,8\cdot46}{50\cdot0,8}\cdot100\%=92\%=92^o\)
\(nC_2H_5OH=\dfrac{2,9}{46}=0,06\left(mol\right)\)
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
0,06 0,06 0,06 0,03 (mol)
VH2 = 0,03.22,4= 0,672 (l)
V = m /D
=> V rượu etylic = 2,9 / 0,8 = 3,625 (ml)
\(m_{C_2H_5OH\left(nguyên.chất\right)}=\dfrac{0,5.30}{100}=0,15l=150ml\)
\(\rightarrow m_{H_2O}=500-150=350ml\)
\(m_{C_2H_5OH}=150.0,8=120g\)
\(m_{H_2O}=350.1=350g\)
\(\left\{{}\begin{matrix}n_{C_2H_5OH}=\dfrac{120}{46}=2,6mol\\n_{H_2O}=\dfrac{350}{18}=19,44mol\end{matrix}\right.\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
2,6 1,3 ( mol )
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
19,44 9,72 ( mol )
\(V_{H_2}=\left(1,3+9,72\right).22,4=246,848l\)
\(0,5lít=500ml\)
\(m_{C_2H_5OH}=500.0,8=400g\)
\(n_{C_2H_5OH}=\dfrac{400}{46}=8,69mol\)
\(n_{Na}=\dfrac{300}{23}=12,04mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
8,69 < 12,04 ( mol )
8,69 8,69 ( mol )
\(V_{H_2}=8,96.22,4=200,704l\)
nH2 = 85,12 : 22,4 = 3,8 (mol) ; nH2O = VH2O.D = 108 (g) => nH2O = 108/18 = 6 (mol)
PTHH:
2Na + 2C2H5OH → 2C2H5ONa + H2↑
x → 0,5x (mol)
2Na + 2H2O → 2NaOH + H2↑
6 → 3 (mol)
Ta có: nH2 = 0,5x + 3 = 3,8
=> x = 1,6 (mol) = nC2H5OH
mC2H5OH = 1,6.46 = 73,6 (g)
