\(n_{C_2H_6O}=\dfrac{m}{M}=\dfrac{9,2}{12\cdot2+6+16}=0,2\left(mol\right)\\ PTHH:C_2H_6O+3O_2-^{t^o}>2CO_2+3H_2O\)

tỉ lệ             1        :      3      :       2      :     3

n(mol)         0,2----->0,6----------->0,4----->0,6

\(V_{CO_2\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\\ V_{H_2O\left(dktc\right)}=n\cdot22,4=0,6\cdot22,4=13,44\left(l\right)\)

PTHH: \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

Ta có: \(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\) 

\(\Rightarrow n_{O_2}=0,5\left(mol\right)\) \(\Rightarrow V_{O_2}0,5\cdot22,4=11,2\left(l\right)\)

a) \(4P+5O_2\underrightarrow{t\text{°}}P_2O_5\)

b)\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\)

Từ PTHH: \(n_{O_2}=\dfrac{5}{4}n_P=\dfrac{5}{4}.0,4=0,5\left(mol\right)\)

\(\Rightarrow\)\(V_{O_2}=0,5.22,4=11,2\left(l\right)\)

\(n_{CO_2}=\dfrac{8.8}{44}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{9.6}{32}=0.3\left(mol\right)\)

\(2CO+O_2\underrightarrow{t^0}2CO_2\)

\(0.2.......0.1.......0.2\)

\(2H_2+O_2\underrightarrow{t^0}2H_2O\)

\(0.4......0.3-0.1\)

\(\%m_{CO}=\dfrac{0.2\cdot28}{0.2\cdot28+0.4\cdot2}\cdot100\%=87.5\%\)

\(\%m_{H_2}=100-87.5=12.5\%\)

a)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2(1)\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O(2) \)

b)

\(n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)\)

Theo PTHH :

\(n_{CO} = n_{CO_2} = 0,2(mol)\\ n_{O_2(1)} = \dfrac{1}{2}n_{CO_2} = 0,1(mol)\\ n_{H_2} = 2n_{O_2(2)} = 2(0,3-0,1) = 0,4(mol)\)

Vậy :

\(\%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\% = 87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

tân thư

môn j z

\(CO+\dfrac{1}{2}O_2\underrightarrow{to}CO_2\\ H_2+\dfrac{1}{2}O_2\underrightarrow{to}H_2O\\ n_{O_2}=\dfrac{1}{2}.\left(n_{CO}+n_{H_3}\right)=\dfrac{0,3}{2}=0,15\left(mol\right)\\ V=V_{O_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\)

2K+2H2O->2KOH+H2

0,1-----0,1-----0,1---0,05

n K=0,1 mol

=>m KOH=0,1.56=5,6g

=>VH2=0,05.22,4=1,12l

2H2+O2-to>2H2O

0,05--------------0,05

n O2=0,2 mol

=>O2 dư

=>mH2O=0,05.18=0,9g

nCO2=0,2mol

PTHH: 2CO+O2=>2CO2

         0,2<--0,1<---0,2

=> mO2=0,2.32=6,4g

=> khối lượng Oxi phản ứng với H2 là :

9,6-6,4=3,2g

=> nH2O=3,2:32=0,1mol

PTHH: 2H+O2=>H2O

           0,2<-0,1<-0,2

=> mH2=2.0,2=0,4g

mCO =0,2.28=5,6g

=> m hh=5,6+0,4=6g

a. \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH : 2H₂ + O₂ -t^o> 2H₂O

             0,3     0,15     0,3

b. \(m_{O_2}=0,15.32=4,8\left(g\right)\)

c. \(m_{H_2O}=0,3.18=5,4\left(g\right)\)

$n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)$

\(2CO+O_2\xrightarrow[]{t^o}2CO_2\)

0,2       0,1       0,2                 (mol)

$n_{O_2} = \dfrac{9,6}{32} = 0,3(mol)$

\(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

0,4     0,2         0,2              (mol)

\(\%m_{CO}=\dfrac{0,2.44}{0,2.44+0,4.2}.100\%=91,67\%\\ \%m_{H_2}=100\%-91,67\%=8,33\%\)

\(\%n_{CO}=\dfrac{0,2}{0,2+0,4}.100\%=33,33\%\\ \%n_{H_2}=100\%-33,33\%=66,67\%\)