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

b) $n_{C_2H_5OH} = \dfrac{4,6}{46} = 0,1(mol)$

$n_{O_2} = 3n_{C_2H_5OH} = 0,3(mol)$
$V_{O_2} = 0,3.22,4 = 6,72(lít)$

c)

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,2(mol) \Rightarrow V_{CO_2} = 0,2.22,4 = 4,48(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,3(mol) \Rightarrow m_{H_2O} = 0,3.18 = 5,4(gam)$

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

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

  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

Cảm ơn bạn nhiều 

C2H5OH + 3O2 => 2CO2 + 3H2O

nC2H5OH = m/M = 9.2/46 = 0.2 (mol)

Theo phương trình => nCO2 = 0.4 (mol), nH2O = 0.6 (mol)

mH2O = n.M = 0.6 x 18 = 10.8 (g)

VCO2 = 22.4 x 0.4 = 8.96 (l)

nC2H5OH= 0.2 mol

C2H5OH + 3O2 -to-> 2CO2 + 3H2O

0.2_________________0.4____0.6

VCO2= 0.4*22.4=8.96l

mH2O= 0.6*18=10.8g

E cảm ơn ạa

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

b) $n_{C_2H_5OH} = \dfrac{9,2}{46} = 0,2(mol)$

Theo PTHH : 

$n_{CO_2} = 2n_{C_2H_5OH} = 0,4(mol) \Rightarrow V_{CO_2} = 0,4.22,4 = 8,96(lít)$

$n_{H_2O} = 3n_{C_2H_5OH} = 0,6(mol) \Rightarrow m_{H_2O} = 0,6.18 = 10,8(gam)$

c) $CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$
$n_{CaCO_3} = n_{CO_2} = 0,4(mol)$

$m_{CaCO_3} = 0,4.100 = 40(gam)$

Bài 2:

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{H_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\ n_{Fe}=n_{H_2}=0,45\left(mol\right);n_{HCl}=2.0,45=0,9\left(mol\right)\\ a,m_{Fe}=0,45.56=25,2\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,9}{0,15}=6\left(M\right)\)

\(n_{O_2}=\dfrac{8}{32}=0.25\left(mol\right)\)

\(CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

\(0.125....0.25....0.125\)

\(m_{CH_4}=0.125\cdot16=2\left(g\right)\)

\(V_{CO_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(............0.125.....0.125\)

\(m_{CaCO_3}=0.125\cdot100=12.5\left(g\right)\)

a)

CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

b) \(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Gọi số mol CH₄, C₂H₂ là a, b (mol)

=> \(\left\{{}\begin{matrix}a+2b=0,3\\16a+26b=4,2\end{matrix}\right.\) 

=> a = 0,1 (mol); b = 0,1 (mol)

=> \(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{16.0,1}{4,2}.100\%=38,095\%\\\%m_{C_2H_2}=\dfrac{26.0,1}{4,2}.100\%=61,905\%\end{matrix}\right.\)