\(n_{CO2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)

  0,1             0,3         0,2

b) \(m_{C2H5OH}=0,1.46=4,6\left(g\right)\)

c) \(V_{O2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

 Chúc bạn học tốt

\(CH_4+2O_2->CO_2+2H_2O\\ n_{CO_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_{NaOH}=80\cdot0,25:40=0,5mol\Rightarrow Tạo:NaHCO_3\\ CO_2+NaOH->NaHCO_3\\ m_{muối}=84\cdot0,5=42\left(g\right)\)

Hỗn hợp gồm: C2H5OH và CH3COOH

C2H5OH + 3O2 => 2CO2 + 3H2O

CH3COOH + 2O2 => 2CO2 + 2H2O

CO2 + Ca(OH)2 dư => CaCO3 +H2O

nO2 = V/22.4 = 11.2 / 22.4 = 0.5 (mol)

nCaCO3 = m/M = 40/100 = 0.4 (mol)

==> nCO2 = 0.4 (mol)

Gọi số mol của C2H5OH và CH3COOH là x,y (mol)

Ta có: 3x + 2y = 0.5; 2x + 2y = 0.4

==> x = y = 0.1 (mol)

a = 0.1 x 46 + 0.1 x 60 = 4.6 + 6 = 10.6 (g)

\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)

a)C2H6O+3O2to→2CO2+3H2O

0,3-------------0,9------0,6

nC2H6O=\(\dfrac{13,8}{46}\)=0,3mol

nCO2=2nC2H6O=0,6mol

⇒VCO2=0,6×22,4=13,44l

⇒Vkk=0,9×22,4×5=100,8l

PTHH: \(C_2H_5OH+3O_2\xrightarrow[]{t^o}2CO_2+3H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{6,4}{46}=\dfrac{16}{115}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{48}{115}\left(mol\right)\\n_{CO_2}=\dfrac{32}{112}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=\dfrac{48}{115}\cdot22,4\approx9,35\left(l\right)\\V_{CO_2}=\dfrac{32}{112}\cdot22,4\approx6,23\left(l\right)\end{matrix}\right.\)

1.\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{KOH}=CM.Vdd=2.0,46=0,92\left(mol\right)\)

PTHH:\(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

TPU:   0,2          0,92

PU:     0,2           0,4            0,2

SPU:    0             0,52            0,2

\(m_{K_2CO_3}=n.M=0,2.138=27,6\left(g\right)\)

2.\(n_{SO_2}=\dfrac{V}{22,4}=\dfrac{8,69}{22,4}=0,4\left(mol\right)\)

\(n_{NaOH}=CM.Vdd=2,5.0,2=0,5\left(mol\right)\)

PTHH:\(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)

TPU:    0,4         0,5

PU:       0,25       0,5            0,25

SPU:    0,15        0               0,25

\(m_{Na_2SO_3}=n.M=0,25.126=31,5\left(g\right)\)

a)

$C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$

b)

n C2H5OH = 9,2/46 = 0,2(mol)

n CO2 = 2n C2H5OH = 0,4(mol) => m CO2 = 0,4.44 = 17,6 gam

n H2O = 3n C2H5OH = 0,6(mol) => m H2O = 0,6.18 = 10,8 gam

c)

n O2 = 3n C2H5OH = 0,6(mol)

=> V O2 = 0,6.22,4 = 13,44(lít)

=> V không khí = 13,44/20% = 67,2 lít

Theo gt ta có: $n_{C_2H_5OH}=0,2(mol)$

a, $C_2H_5OH+3O_2\rightarrow 2CO_2+3H_2O$

b, Ta có: $n_{CO_2}=0,4(mol)\Rightarrow m_{CO_2}=17,6(g)$

$n_{H_2O}=0,6(mol)\Rightarrow m_{H_2O}=10,8(g)$

c, Ta có: $n_{O_2}=0,6(mol)\Rightarrow V_{O_2}=13,44(l)\Rightarrow V_{kk}=67,2(l)$

a, nC2H4 = 2,24/22,4 = 0,1 (mol)

PTHH: C2H4 + 3O2 -to-> 2CO2 + 2H2O

Mol: 0,1 ---> 0,3 ---> 0,2 

b, VO2 = 0,3 . 22,4 = 6,72 (l)

c, mCO2 = 0,2 . 44 = 8,8 (g)

d, Vkk = 6,72 . 5 = 33,6 (l)

PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O

Mol: 0,2 <--- 0,2 ---> 0,2

mCaCO3 = 0,2 . 100 = 20 (g)

a, \(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)

Ta có: \(n_{Br_2}=\dfrac{64}{160}=0,4\left(mol\right)\)

Theo PT: \(n_{C_2H_2}=\dfrac{1}{2}n_{Br_2}=0,2\left(mol\right)\)

Mà: \(n_X=\dfrac{10,08}{22,4}=0,45\left(mol\right)\Rightarrow n_{CH_4}=0,25\left(mol\right)\)

\(\Rightarrow V_{CH_4}=0,25.22,4=5,6\left(l\right)\)

c, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_2}=0,65\left(mol\right)\)

PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)

Theo PT: \(n_{CaCO_3}=n_{CO_2}=0,65\left(mol\right)\Rightarrow m_{CaCO_3}=0,65.100=65\left(g\right)\)