a) \(n_{CH_4}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)

b) Theo PTHH: \(n_{H_2O}=n_{O_2}=2n_{CH_4}=2.0,4=0,8\left(mol\right)\)

\(m_{H_2O}=n.M=0,8.18=14,4\left(g\right)\)

c) \(V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,8.22,4=17,92\left(l\right)\)

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 

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

nCH4 = 3,36 : 22,4 = 0,15 (mol) 
pthh : CH4 + 2O2 -t--> CO2 + 2H2O 
         0,15      0,3          
=> VO2 = 0,3 . 22,4 = 6,72 (L) 
ta có  : VO2 = 20% Vkk => Vkk = VO2 : 20% = 6,72 : 20% = 33,6 (L)

\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:0,15\rightarrow0,3\\ \rightarrow\left\{{}\begin{matrix}V_{O_2}=0,3.22,4=6,72\left(l\right)\\V_{kk}=6,72.5=33,6\left(l\right)\end{matrix}\right.\)

\(n_{CH_4}=\dfrac{2,24}{22,4}=0,1mol\)

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

0,1         0,2    0,1

\(V_{O_2}=0,2\cdot22,4=4,48l\)

\(V_{CO_2}=0,1\cdot22,4=2,24l\)

\(n_{CH_4}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

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

\(0.05.....0.1.......0.05\)

\(V_{O_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(V_{CO_2}=0.05\cdot22.4=1.12\left(l\right)\)

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

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

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

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

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

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

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

$n_{C_2H_2} = \dfrac{6,72}{22,4} = 0,3(mol) ; n_{O_2} = 0,5(mol)$
$2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O$

Ta thấy : 

$n_{C_2H_2} : 2 > n_{O_2} : 5$ nên $C_2H_2$ dư

Theo PTHH : 

$n_{C_2H_2\ pư} = \dfrac{5}{2} = 0,2(mol)$
$n_{CO_2} = 0,4(mol) ; n_{H_2O} = 0,2(mol)$

Suy ra : 

$m_{C_2H_2\ dư} = (0,3 - 0,2).26 = 2,6(gam)$
$m_{CO_2} = 0,4.44 = 17,6(gam)$
$m_{H_2O} = 0,2.18 = 3,6(gam)$

Bài 1:

a, PT: \(Na_2O+H_2O\rightarrow2NaOH\)

b, Ta có: \(n_{Na_2O}=\dfrac{31}{62}=0,5\left(mol\right)\)

\(n_{H_2O}=\dfrac{27}{18}=1,5\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,5}{1}< \dfrac{1,5}{1}\), ta được H₂O dư.

Theo PT: \(n_{NaOH}=2n_{Na_2O}=1\left(mol\right)\)

\(\Rightarrow m_{NaOH}=1.40=40\left(g\right)\)

b, Theo PT: \(n_{H_2O\left(pư\right)}=n_{Na_2O}=0,5\left(mol\right)\)

\(\Rightarrow n_{H_2O\left(dư\right)}=1,5-0,5=1\left(mol\right)\)

\(\Rightarrow m_{H_2O\left(dư\right)}=1.18=18\left(g\right)\)

Bài 2:

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

Ta có: \(n_{CH_4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

b, Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,15}{2}\), ta được CH₄ dư.

Theo PT: \(n_{CH_4\left(pư\right)}=\dfrac{1}{2}n_{O_2}=0,075\left(mol\right)\)

\(\Rightarrow n_{CH_4\left(dư\right)}=0,1-0,075=0,025\left(mol\right)\)

\(\Rightarrow V_{CH_4\left(dư\right)}=0,025.22,4=0,56\left(l\right)\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{1}{2}n_{O_2}=0,075\left(mol\right)\\n_{H_2O}=n_{O_2}=0,15\left(mol\right)\end{matrix}\right.\)

⇒ m _{sản phẩm} = m_CO2 + m_H2O = 0,075.44 + 0,15.18 = 6 (g)