Ta có: \(n_{CH_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

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

____0,2___0,4 (mol)

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

Bạn tham khảo nhé!

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

PTHH: CH4 + 2O2 -> (t°) CO2 + 2H2O

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

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

Câu 1:

\(a,PTHH:C_2H_4+5O_2\underrightarrow{t^o}2CO_2+2H_2O\)

\(n_{C_2H_4}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(Theo.PTHH:n_{O_2}=5.n_{C_2H_4}=5.0,25=1,25\left(mol\right)\\ V_{O_2\left(đktc\right)}=n.22,4=1,25.22,4=28\left(l\right)\)

\(b,\Rightarrow V_{kk\left(đktc\right)}=5.V_{O_2\left(đktc\right)}=5.28=140\left(l\right)\)

B1:

\(C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\)

Vì số mol tỉ lệ thuận với thể tích, đồng thời nhìn PTHH, ta sẽ được:

\(a,V_{O_2\left(đktc\right)}=3.V_{C_2H_4\left(đktc\right)}=3.5,6=16,8\left(l\right)\)

\(b,V_{kk}=5.V_{O_2\left(đktc\right)}=16,8.5=84\left(l\right)\)

B2:

Đặt số mol metan, etylen lần lượt là a,b (mol) (a,b>0)

\(n_{hh}=n_{CH_4}+n_{C_2H_4}=a+b=\dfrac{3,36}{22,4}=0,15\left(1\right)\)

PTHH: CH₄ +2 O₂ -t^o-> CO₂ +2 H₂O

C₂H₄ +3 O₂ -t^o-> 2CO₂ + 2H2O

\(n_{CO_2\left(tổng\right)}=a+2b=\dfrac{8,8}{44}=0,2\left(mol\right)\left(2\right)\)

(1), (2) =>a=0,1; b=0,05

Số mol tỉ lệ tương ứng với thể tích. Nên:

\(\%V_{CH_4}=\%n_{CH_4}=\dfrac{0,1}{0,15}.100\approx66,667\%\\ \Rightarrow\%V_{C_2H_4}\approx33,333\%\)

\(a,PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ n_{Al}=\dfrac{m}{M}=\dfrac{10,8}{27}=0,4\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,4=0,2\left(mol\right)\\ m_{Al_2O_3}=n.M=0,2.102=20,4\left(g\right)\)

\(b,n_{O_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ Lập.tỉ.lệ:\dfrac{n_{Al}}{4}>\dfrac{n_{O_2}}{3}\Rightarrow Al.dư\\ Theo.PTHH:n_{Al\left(pư\right)}=\dfrac{4}{3}.n_{O_2}=\dfrac{4}{3}.0,2\left(mol\right)\\ n_{Al\left(dư\right)}=n_{Al\left(bđ\right)}-n_{Al\left(pư\right)}=0,4-0,2=0,2\left(mol\right)\\ Theo.PTHH:n_{Al_2O_3}=\dfrac{1}{2}.n_{Al}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\\ m_{Al_2O_3}=n.M=0,1=102=10,2\left(g\right)\)

a)

\(n_{O_2} = \dfrac{11,2}{22,4} = 0,5(mol)\\ 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ n_P = \dfrac{4}{5}n_{O_2} = 0,4(mol)\\ \Rightarrow m_P = 0,4.31 = 12,4(gam)\)

b)

\(n_{P_2O_5} = \dfrac{2}{5}n_{O_2} = 0,2(mol)\\ \Rightarrow m_{P_2O_5} = 0,2.142 = 28,4(gam)\)

c)

\(2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{KMnO_4} = 2n_{O_2} = 0,5.2 = 1(mol)\\ \Rightarrow m_{KMnO_4} = 1.158 = 158(gam)\)

a)

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

\(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

            0,1<-0,05<-------0,1

            2CO + O₂ --t^o--> 2CO₂

            0,2<--0,1-------->0,2

=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\V_{CO}=0,2.22,4=4,48\left(l\right)\end{matrix}\right.\)

b) \(m_{CO_2}=0,2.44=8,8\left(g\right)\)

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

\(n_{H_2O}=\dfrac{1,8}{18}=0,1mol\)

\(2CO+O_2\rightarrow2CO_2\)

  a         0,5a     a

\(2H_2+O_2\rightarrow2H_2O\)

 0,1   0,05 \(\leftarrow\)    0,1

\(\Sigma n_{O_2}=0,5a+0,05=0,15\)

\(\Rightarrow a=n_{O_2\left(CO\right)}=0,2mol\)

\(V_{CO}=2\cdot0,2\cdot22,4=8,96l\)

\(V_{H_2}=0,1\cdot22,4=2,24l\)

\(m_{CO_2}=0,2\cdot44=8,8g\)

\(1) n_{KMnO_4}= \dfrac{31,6}{158} = 0,2(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ n_{O_2} = \dfrac{1}{2}n_{KMnO_4} = 0,1(mol)\\ V_{O_2} = 0,1.22,4 = 2,24(lít)\\ 2) CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O\\ V_{CH_4} = \dfrac{1}{2}V_{O_2} = 1,12(lít)\\ 3)n_{CH_4} = \dfrac{1,12}{22,4} = 0,05(mol)\\ \text{Nhiệt lượng tỏa ra = } = 0,05.880 = 44(KJ)\)

  Sry chữ hơi xấu 🙂

a.\(n_{CH_4}=\dfrac{V_{CH_4}}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)

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

0,3       0,6                                          ( mol )

\(V_{O_2}=n_{O_2}.22,4=0,6.22,4=13,44l\)

b.

\(n_P=\dfrac{m_P}{M_P}=\dfrac{3,1}{31}=0,1mol\)

\(4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

0,1                         0,05    ( mol )

\(m_{P_2O_5}=n_{P_2O_5}.M_{P_2O_5}=0,05.142=7,1g\)

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)

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