43. \(4Al+3O_2\xrightarrow{t^o}2Al_2O_3\)

\(\Rightarrow n_{O_2}=\dfrac{3}{4}n_{Al}=\dfrac{3}{4}.\dfrac{10,8}{27}=0,3(mol)\\ \Rightarrow V_{O_2}=0,3.22,4=6,72(l)\\ \Rightarrow D\)

44. KL ko tan là Cu 

\(Fe+H_2SO_4\to FeSO_4+H_2\)

\(\Rightarrow n_{Fe}=n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow m=m_{Cu}=m_{hh}-m_{Fe}=10-0,1.56=4,4(g)\)

Chọn C

\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\\ n_{O_2}=\dfrac{n_{H_2}}{2}\Rightarrow V_{O_2\left(đktc\right)}=\dfrac{V_{H_{2\left(đktc\right)}}}{2}=\dfrac{2,24}{2}=1,12\left(l\right)\\ Chọn.A\)

\(n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,2\rightarrow0,6\rightarrow0,4\rightarrow0,4\\ \rightarrow\left\{{}\begin{matrix}V_{kk}=0,6.5.22,4=67,2\left(l\right)\\V_{CO_2}=0,4.44=17,6\left(g\right)\\m_{H_2O}=0,4.18=7,2\left(g\right)\end{matrix}\right.\)

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH:

2CO + O₂ --t^o--> 2CO₂

0,2<---0,1<--------0,2

2H₂ + O₂ --t^o--> 2H₂O

0,4<--0,2<-------0,2

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\\%V_{H_2}=100\%-33,33\%=66,67\%\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: 2CO + O₂ --t^o--> 2CO₂

            0,3<--0,15<------0,3

            2H₂ + O₂ --t^o--> 2H₂O

            0,1<--0,05

\(\Rightarrow\left\{{}\begin{matrix}\%V_{CO}=\%n_{CO}=\dfrac{0,3}{0,3+0,1}.100\%=75\%\\\%V_{H_2}=100\%-75\%=25\%\end{matrix}\right.\)

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)

\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15mol\)

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

0,15     0,3

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

Không có đáp án đúng!

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PT: \(S+O_2\underrightarrow{t^o}SO_2\)

Theo PT: \(n_{SO_2}=n_{O_2}=0,3\left(mol\right)\Rightarrow V_{SO_2}=0,3.22,4=6,72\left(l\right)\)

Mọi người giải luôn hộ nhé

CH4+2O2-to>CO2+2H2O

0,5-----1 mol

n CH4=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>Vkk=1.22,4.5=112l

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