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![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{C_6H_{12}O_6}=\dfrac{m}{M}=\dfrac{1,8}{12\cdot6+12+16\cdot6}=0,01\left(mol\right)\\ PTHH:C_6H_{12}O_6+6O_2-^{t^o}>6CO_2+6H_2O\)
tỉ lệ: 1 : 6 : 6 : 6
n(mol) 0,01------->0,06------->0,06------>0,06
\(V_{CO_2\left(dktc\right)}=n\cdot22,4=0,08\cdot22,4=1,792\left(l\right)\) khí CO2 là đo ở điều kiện nào nhỉ?
\(m_{H_2O}=n\cdot M=0,06\cdot18=1,08\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi \(\left\{{}\begin{matrix}n_{CH_4}=x\\n_{H_2}=y\end{matrix}\right.\)
\(n_{O_2}=\dfrac{7,437}{24,79}=0,3mol\)
\(n_{H_2O}=\dfrac{7,2}{18}=0,4mol\)
\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)
x 2x x 2x ( mol )
\(2H_2+O_2\rightarrow\left(t^o\right)2H_2O\)
y 1/2 y y ( mol )
Ta có:
\(\left\{{}\begin{matrix}2x+\dfrac{1}{2}y=0,3\\2x+y=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(m=m_{CH_4}+m_{H_2}=\left(0,1.16\right)+\left(0,2.2\right)=2g\)
\(V_{CO_2}=0,1.24,79=2,479l\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Số mol khí CH4 là: nCH4= =2 mol
PTHH: CH4 + 2O2 CO2 + 2H2O
Tỉ lệ PT: 1mol 2mol
Phản ứng: 2mol → 4mol
=> thể tích khí oxi cần dùng là: VO2=22,4.n=22,4.4=89,6 lít
Đáp án cần chọn là: B
\(n_{CH_4}=\dfrac{32}{16}=2\left(mol\right)\)
PTHH: CH4 + 2O2 --to--> CO2 + 2H2O
2---->4
=> V = 4.22,4 = 89,6 (l)
=> B
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Theo giả thiết ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)
\(4P+5O_2--t^o->2P_2O_5\)
Ta có: \(n_{O_2}=\dfrac{5}{4}.n_P=0,125\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=0,125.22,4=2,8\left(l\right)\)
b, Theo giả thiết ta có: \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(CH_4+2O_2--t^o->CO_2+2H_2O\)
Ta có: \(n_{O_2}=2.n_{CH_4}=0,1\left(mol\right)\Rightarrow V_{O_2\left(đktc\right)}=2,24\left(l\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{O_2}=\dfrac{44,8}{22,4}.20\%=0,4(mol)\)
Bảo toàn NT (O): \(n_{O_2}=n_{CO_2}=\dfrac{1}{2}n_{H_2O}\)
\(\Rightarrow n_{CO_2}=0,4(mol);n_{H_2O}=0,8(mol)\\ \Rightarrow V_{CO_2}=0,4.22,4=8,96(g);m_{H_2O}=0,8.18=14,4(g)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{O_2}=\dfrac{41,44}{22,4}.20\%=0,37(mol)\\ n_{H_2O}=\dfrac{4,68}{18}=0,26(mol)\)
Bảo toàn nguyên tố (O): \(n_{CO_2}=n_{O_2}=0,37(mol)\)
\(\Rightarrow V_{CO_2}=0,37.22,4=8,288(l)\)
BTKL: \(m_{hh}=m_{CO_2}+m_{H_2O}-m_{O_2}=0,37.44+4,68-0,37.32=9,12(g)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có PT: S + O2 ---> SO2. (1)
nS = 1,6/32=0,05(mol)
Theo PT(1), ta có:
nO2=nS=0,05(mol)
=> V=0,05.22,4=1,12(l)
2KNO3 ---> 2KNO2 + O2. (2)
Theo PT(2), ta có: nKNO3=2.nO2=2.0,05=0,1(mol)
=> m = 101.0,1=10,1(g)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2S\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\ PTHH:2H_2S+3O_2-^{t^o}>2SO_2+2H_2O\)
tỉ lệ: 2 : 3 : 2 : 2
n(mol) 0,5---->0,75------>0,5------->0,5
\(m_{SO_2}=n\cdot M=0,5\cdot64=32\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{CO_2}=\dfrac{13,2}{44}=0,3\left(mol\right)\)
=> nC = 0,3 (mol)
\(n_{H_2O}=\dfrac{4,32}{18}=0,24\left(mol\right)\)
=> nH = 0,48 (mol)
m = mC + mH = 0,3.12 + 0,48.1 = 4,08 (g)
\(n_{CH_4}=\dfrac{1,6}{16}=0,1\left(mol\right)\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\)
0,1--->0,2----->0,1---->0,2
\(\Rightarrow\left\{{}\begin{matrix}V=V_{CO_2}=0,1.22,4=2,24\left(l\right)\\m=m_{H_2O}=0,2.18=3,6\left(g\right)\end{matrix}\right.\)