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\(pthh:C+O_2\overset{t^o}{--->}CO_2\left(1\right)\)
\(S+O_2\overset{t^o}{--->}SO_2\left(2\right)\)
Ta có: \(\left\{{}\begin{matrix}n_C=\dfrac{6}{12}=0,5\left(mol\right)\\n_S=\dfrac{8}{32}=0,25\left(mol\right)\end{matrix}\right.\)
Theo pt(1): \(n_{O_2}=n_C=0,5\left(mol\right)\)
Theo pt(2): \(n_{O_2}=n_S=0,25\left(mol\right)\)
\(\Rightarrow m_{O_{2_{cần.dùng}}}=0,5.32+0,25.32=24\left(g\right)\)
Chọn B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{O_2}=\dfrac{20-15,2}{32}=0,15\left(mol\right)\)
=> V = 0,15.22,4 = 3,36 (l)
=> D
![](https://rs.olm.vn/images/avt/0.png?1311)
a) C + O2 --to--> CO2
b) \(n_C=\dfrac{6}{12}=0,5\left(mol\right)\)
PTHH: C + O2 --to--> CO2
0,5-->0,5------->0,5
=> mCO2 = 0,5.44 = 22 (g)
c) VO2 = 0,5.22,4 = 11,2 (l)
d) Vkk = 11,2.5 = 56 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_C=\dfrac{2,4}{12}=0,2\left(mol\right)\\ n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\\ PTHH:C+O_2\underrightarrow{t^o}CO_2\\ LTL:0,2>0,15\Rightarrow C.du\\ Theo.pt:n_{CO_2}=n_{O_2}=0,15\left(mol\right)\\ m_{CO_2}=0,15.44=6,6\left(g\right)\)
Số mol khí cacbon và khí oxi lần lượt là 2,4/12=0,2 (mol) và 4,8/32=0,15 (mol).
Khối lượng khí CO2 lớn nhất thu được là 0,15.44=6,6 (g).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_C=\dfrac{14,4}{44}=\dfrac{18}{55}\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ n_{O_2}=n_C=n_{CO_2}=\dfrac{18}{55}\left(mol\right)\\ a,V_{O_2\left(đktc\right)}=\dfrac{18}{55}.22,4=\dfrac{2016}{275}\left(lít\right)\\ b,V_{kk}=\dfrac{100}{21}.\dfrac{2016}{275}=\dfrac{381}{11}\left(lít\right)\\ c,2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3\left(LT\right)}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.\dfrac{18}{55}=\dfrac{12}{55}\left(mol\right)\\ \Rightarrow n_{KClO_3\left(TT\right)}=120\%.\dfrac{12}{55}=\dfrac{72}{275}\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.\dfrac{72}{275}=\dfrac{1764}{55}\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
4Al+3O2-to>2Al2O3
0,04---0,03------0,02 mol
n Al=\(\dfrac{1,08}{27}\)=0,04 mol
=>VO2=0,03.22,4=0,672l
b)
2A+O2-to>2AO
0,06--0,03 mol
=>\(\dfrac{3,84}{A}=0,06\)
=>A=64 :=>Al là Đồng
![](https://rs.olm.vn/images/avt/0.png?1311)
nAl=16,2/27= 0,6(mol)
a) PTHH: 4 Al +3 O2 -to-> 2 Al2O3
nO2= 3/4 . nAl=3/4 . 0,6= 0,45(mol)
=> V(O2,đktc)=0,45 x 22,4=10,08(l)
b) nAl2O3= nAl/2=0,6/2=0,3(mol)
=>mAl2O3=102. 0,3= 30,6(g)
c) 2KMnO4 -to-> K2MnO4 + MnO2 + O2
nKMnO4= 2.nO2=2. 0,45=0,9(mol)
=>mKMnO4= 158 x 0,9= 142,2(g)