\(V_{O_2}=0,125.22,4=2,8\left(l\right)\)

\(n_{H_2}=\dfrac{3}{2}=1,5mol\)

\(V_{O_2}=2.22,4=44,8l\)

TK

Từ C₂H₄O₂ ta có: M = 60 g/mol; m_C = 2 x 12 = 24 g; m_H = 4 x 1 = 4 g;

M_O = 2 x 16 = 32 g.

%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;

%O = 100% - 40% - 6,67% = 53,33%.

nC2H2= 0,2(mol)

PTHH: 2 C2H2 + 5 O2 -to-> 4 CO2 + 2 H2O 

nO2= 5/2 x 0,2=0,5(mol)

=> V(O2,ddktc)=0,5.22,4=11,2(l)

pt: \(2C_2H_2+5O_2\rightarrow4CO_2+2H_2O\)

Theo pt: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=\dfrac{5}{2}.\dfrac{5,2}{26}=0,5mol\)

\(\Rightarrow V_{O_2}=0,5.22.4=11,2lit\)

\(n_{C_2H_2} = \dfrac{5,2}{26} = 0,2(mol)\\ C_2H_2 + \dfrac{5}{2}O_2 \xrightarrow{t^o} 2CO_2 + H_2O\\ n_{O_2} = \dfrac{5}{2}n_{C_2H_2} = 0,5(mol\\ \Rightarrow V_{O_2} = 0,5.22,4 = 11,2(lít)\)

\(V_{O_2}=n_{O_2}\times22,4=0,15\times22,4=3,36\left(l\right)\)

V_O2(đktc) = 0,15 x 22,4 = 3,36 lít

Chúc bạn học tốt!!

nếu mà muốn tính khối lượng 0,15 mol cu thì thế nào ạ?

16 gam khí oxi: n = m : M = 16 : 32 = 0,5 (mol)

V = n.22,4 = 0,5.22,4 = 11,2 (lít)

CH4+2O2-to>CO2+2H2O

0,075---0,15

n O2=\(\dfrac{3,36}{22,4}\)=0,15 mol

=>VCH4=0,075.22,4=1,68l

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

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

 0,075  0,15                                    ( mol )

\(V_{CH_4}=0,075.22,4=1,68l\)

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

   CH₄ + 2O₂ \(\underrightarrow{t^o}\) CO₂ + 2H₂O

  0,075    0,15                               ( mol )

\(V_{CH_4}=0,075.22,4=1,68l\)