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

\(n_{C_2H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(C_2H_2+\dfrac{5}{2}O_2\underrightarrow{^{t^0}}2CO_2+H_2O\)

\(0.1........0.25...........0.2\)

\(V_{O_2}=0.25\cdot22.4=5.6\left(l\right)\)

\(V_{CO_2}=0.2\cdot22.4=4.48\left(l\right)\)

n C2H2 = 22,4/22,4 = 1(mol)

$2C_2H_2 + 5O_2 \xrightarrow{t^o} 4CO_2 + 2H_2O$

Theo PTHH :

n O2 = 5/2 n C2H2 = 2,5(mol) => V O2 = 2,5.22,4 = 56(lít)

n CO2 = 2n C2H2 = 2(mol) => V CO2 = 2.22,4 = 44,8(lít)

\(V_{O_2}=\dfrac{336}{5}=67,2\left(ml\right)=0,0672\left(l\right)\\ n_{O_2}=\dfrac{0,0672}{22,4}=0,003\left(mol\right)\\ CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\\ n_{CO_2}=n_{CH_4}=\dfrac{0,003}{2}=0,0015\left(mol\right)\\ a,V_{CH_4\left(đktc\right)}=0,0015.22,4=0,0336\left(l\right)\\ b,V_{CO_2\left(đktc\right)}=V_{CH_4\left(đktc\right)}=0,0336\left(l\right)\)

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

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,2--->0,4--------->0,2

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

a)2H₂ + O₂ --t^o--> 2H₂O

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,5-->0,25------>0,5

=> \(m_{H_2O}=0,5.18=9\left(g\right)\)

c) V_O2 = 0,25.22,4 = 5,6 (l)

=> V_kk = 5,6 : 20% = 28 (l)

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

PTHH: CH₄ + 2O₂ ---t^o→ CO₂ + 2H₂O

Mol:      0,2      0,4

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

\(CH_4+ZO_2\rightarrow CO_2+2H_2O\)

0,2         0,4        0,2

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

\(\Rightarrow V_{O_2}=8,96l\\ m_{CO_2}=n.M=0,2.44=8,8\left(g\right)\)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

_____0,1-->0,125---->0,05

=> m_P2O5 = 0,05.142 = 7,1 (g)

c) V_O2 = 0,125.22,4 = 2,8(l)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b)0,1mol

c, 2,8l

\(n_{CH_4}=\dfrac{3.2}{22.4}=\dfrac{1}{7}\left(mol\right)\)

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

\(\dfrac{1}{7}.............\dfrac{1}{7}\)

\(V_{CO_2}=\dfrac{1}{7}\cdot22.4=3.2\left(l\right)\)