\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\)

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)

0,6       0,3                  0,6        ( mol )

\(m_{H_2O}=0,6.18=10,8g\)

\(V_{kk}=V_{O_2}.5=\left(0,3.22,4\right).5=33,6l\)

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

Ta có: \(n_{CH_4}=\dfrac{30,9875}{24,79}=1,25\left(mol\right)\)

a, \(n_{H_2O}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow m_{H_2O}=2,5.18=45\left(g\right)\)

b, \(n_{O_2}=2n_{CH_4}=2,5\left(mol\right)\) \(\Rightarrow V_{O_2}=2,5.24,79=61,975\left(l\right)\)

Mà: O₂ chiếm 1/5 thể tích không khí.

\(\Rightarrow V_{kk}=5V_{O_2}=309,875\left(l\right)\)

\(n_{C_2H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ a,2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ b,n_{CO_2}=0,125.2=0,25\left(mol\right)\\ m_{CO_2}=0,25.44=11\left(g\right)\\ c,n_{O_2}=\dfrac{5}{2}.0,125=0,3125\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,3125.22,4=7\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=\dfrac{100}{20}.7=35\left(l\right)\)

\(n_{CH_4}=\dfrac{0.8}{16}=0.05\left(mol\right)\)

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

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

\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)

\(m_{H_2O}=0.1\cdot18=1.8\left(g\right)\)

\(n_{C_2H_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

C₂H₄ + 3O₂ ----t^o---> 2CO₂ + 2H₂O

0,4        1,2                                0,8

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

\(V_{kk}=5V_{O_2}=5.1,2.22,4=134,4\left(l\right)\)

\(n_{C_2H_6}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: \(2C_2H_6+7O_2\xrightarrow[]{t^o}4CO_2+6H_2O\)

             0,25--->0,875

`=> V_{O_2} = (0,875.22,4)/(20%) = 98 (l)`

2H2+O2-to>2H2O

0,1----0,05----0,1

n H2=0,1 mol

=>m H2OI=0,1.18=1,8g

=>Vkk=0,05.22,4.5=5,6l

câu này là câu a hay b z bn ?

C₂H₂+\(\dfrac{5}{2}\)O₂-to>2CO₂+H₂O

0,05--0,125---------0,1 mol

n _C2H2=\(\dfrac{1,12}{22,4}\)=0,05 mol

=>m _CO2=0,1.44=4,4g

=>Vkk=0,125.5.22,4=5,6l