nCH4 = 4,48/22,4 = 0,2 (mol)

PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O

Mol: 0,2 ---> 0,4 ---> 0,2 ---> 0,4

VO2 = 0,4 . 22,4 = 8,96 (l)

VCO2 = 0,2 . 22,4 = 4,48 (l)

Vkk = 8,96 . 5 = 44,8 (l)

a. \(n_{CH_4}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

PTHH : CH₄ + 2O₂ ---t^o---> CO₂ + 2H₂O

           0,2        0,4                0,2       

b. \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c. \(V_{kk}=8,96.5=44,8\left(l\right)\)

a. \(n_{CH_4}=\dfrac{4.48}{22,4}=0,2\left(mol\right)\)

PTHH : CH₄ + 2O₂ ---t⁰---> CO₂ + 2H₂O

            0,2       0,4               0,2

b. \(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c. \(V_{kk}=8,96.5=44,8\left(l\right)\)

Bài 1:

PTHH: \(CH_4+Cl_2\underrightarrow{a/s}CH_3Cl+HCl\)

Theo PTHH: \(n_{Cl_2}=n_{CH_3Cl}=n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CH_3Cl}=0,15\cdot50,5=7,575\left(g\right)\\V_{CH_4}=V_{Cl_2}=3,36\left(l\right)\end{matrix}\right.\) 

Bài 2:

PTHH: \(CH_4+2O_2\underrightarrow{a/s}CO_2+2H_2O\)

Theo PTHH: \(\left\{{}\begin{matrix}n_{CO_2}=n_{CH_4}\\n_{O_2}=2n_{CH_4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=V_{CH_4}=5,6\left(l\right)\\V_{CO_2}=2V_{CH_4}=11,2\left(l\right)\end{matrix}\right.\)

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

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

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

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

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

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

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

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

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ pthh:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
               0,2    0,4                  0,4 
\(V_{O_2}=0,4.22,4=8,96\left(l\right)\\ m_{H_2O}=0,4.18=7,2\left(g\right)\)

a) PTHH: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b+c)

Vì trong chất khí, tỉ lệ số mol cũng chính là tỉ lệ về thể tích 

\(\Rightarrow\left\{{}\begin{matrix}V_{CO_2}=V_{CH_4}=11,2\left(l\right)\\V_{O_2}=2V_{CH_4}=22,4\left(l\right)\end{matrix}\right.\)

\(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,2--->0,4--------->0,2

\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=0,4.22,4=8,96\left(l\right)\\m_{CO_2}=0,2.44=8,8\left(g\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\)

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

\(0.1.......0.2........0.1..........0.2\)

\(m_{CH_4}=0.1\cdot16=1.6\left(g\right)\)

\(V_{H_2O}=0.2\cdot22.4=4.48\left(l\right)\)

\(V_{kk}=5V_{O_2}=5\cdot0.2\cdot22.4=22.4\left(l\right)\)

a)

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

b) Theo PTHH : $V_{O_2} = 2V_{CH_4} = 11,2(lít)$

$n_{CH_4} = \dfrac{5,6}{22,4} = 0,25(mol)$

Theo PTHH : $n_{H_2O} = 2n_{CH_4} = 0,5(mol)$
$m_{H_2O} = 0,5.18 = 9(gam)$

\(a,CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(1:2:1:2\left(mol\right)\)

\(0,25:0,5:0,25:0,5\left(mol\right)\)

\(n_{CH_4}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)

\(m_{H_2O}=n.M=0,5.18=9\left(g\right)\)