a) PTHH: C + O2 -to-> CO2

x_____________x_____x(mol)

S+ O2 -to-> SO2

y__y________y(mol)

b) Ta có: 

\(\left\{{}\begin{matrix}12x+32y=5,6\\32x+32y=9,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

mC=0,2.12=2,4(g)

mS=0,1.32=3,2(g)

c) 

\(\%mC=\dfrac{2,4}{5,6}.100\approx42,857\%\\ \rightarrow\%mS\approx100\%-42,857\%\approx57,143\%\)

d)

 \(\%nCO2=\dfrac{x}{x+y}.100\%=\dfrac{0,2}{0,2+0,1}.100\approx66,667\%\\ \rightarrow\%nSO2=\dfrac{y}{x+y}.100\%=\dfrac{0,1}{0,2+0,1}.100\approx33,333\%\)

a)nO2=m/M=9,6/32=0,3 (mol)

C + O2 ->t° CO2

1:1:1

x/12 :(x/12) :x/12 mol

S + O2->t° SO2

1:1:1

5,6-x/32: (5,6-x/32): 5,6-x/32 mol

gọi x là số gam của cacbon 

nC=m/M=x/12(mol)

nS=5,6-x/12 (mol)

b)ta có phương trinh

5,6-x/32+x/12=0,3

<=>3(5,6-x)/96 + 8x/96= 28,8/96

->3(5,6-x)+8x=28,8

<=> 16,8 -3x+8x=28,8

<=>-3x+8x=12

<=>5x=12

<=>x=2,4

-> mC=2,4(g)

mS=5,6-2,4=3,2(g)

c)%mC=2,4/5,6.100%= 42,857%

%mS=100%-42,857%=57,143%

d)%nCO2=0,2/0,3.100%=66,7%

%nSO2=100%-66,7%=33,3%

A/ nO2=0,3 mol

C   +       O2----->    Co2

x mol       x mol          xmol

S+           O2------>  SO2

y mol      y mol          y mol

Ta co     x+y=0,3

12x+32y=5,6

=> x=0,2     y=0,1

B/mC=0,2.12=2,4g    mS= 0,1.32=3,2g

C/ %mC=(2,4/5,6).100=42,8%

     %mS=57,2%

D/ %Co2=(0,2/0,3).100=66,7%

     %So2=33,3%

nO2=0,3mol

gọi x,y là số mol của C và S trong hh

PTHH: C+O2=>CO2

            x->x------x>

            S+O2=>SO2

            y->y------>y

theo 2 pthh trên ta có hpt: 

\(\begin{cases}12x+32y=5,6\\x+y=0,3\end{cases}\)

<=> \(\begin{cases}x=0,2\\y=0,1\end{cases}\)

=> mC=0,2.12=2,4g

=> mS=5,6-2,4=3,2g

%mC=2,4/5,6.100=41,89%

=>%mO=100-41,89=58,11%

m khí thu được =mCO2+SO2=0,2.44+0,1.64=15,2g

=> %mCO2=0,2.44/15,2.100=57,89%

=>%mSO2=100-57,89=42,11%

PTHH: 2CO+O2to→2CO2 (1)

                4H2+O2to→2H2O  (2)

b) Ta có:

ΣnO2=\(\dfrac{9,6}{32}\)=0,3(mol)

nCO2=\(\dfrac{8,8}{44}\)=0,2(mol)

⇒{nO2(1)=0,1mol

nO2(2)=0,2mol

⇒{mCO=0,1⋅28=2,8(g)

mH2=0,2⋅2=0,4(g)

 ⇒%mCO=\(\dfrac{2,8}{2,8+0,4}\)⋅100%=87,5%

%mH2=12,5%

\(nO_2=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(nCO_2=\dfrac{8,8}{44}=0,2\left(mol\right)\)

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

  2        1         2   (mol)

0,2       0,1      0,2   (mol)

\(4H_2+O_2\underrightarrow{t^o}2H_2O\)

4         1         2     (mol)

0,8       0,2      0,4 (mol)

\(mCO=0,2.28=5,6\left(g\right)\)

\(mH_2=0,8.2=0,16\left(g\right)\)

\(\%mCO=\dfrac{5,6.100}{5,6+0,16}=97,22\%\)

\(\%mH_2=100-97,22=2,78\%\)

\(2CO + O_2 \xrightarrow{t^o} 2CO_2\\ 2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{CO} = n_{CO_2} = \dfrac{8,8}{44} = 0,2(mol)\\ n_{O_2} = \dfrac{n_{CO} + n_{H_2}}{2}=\dfrac{0,2+n_{H_2}}{2} = \dfrac{9,6}{32} = 0,3(mol)\\ \Rightarrow n_{H_2} = 0,4(mol)\\ \%V_{CO} = \dfrac{0,2}{0,2 + 0,4}.100\% = 33,33\%\\ \%V_{H_2} = 100\% - 33,33\% = 66,67\%\\ \%m_{CO} = \dfrac{0,2.28}{0,2.28+0,4.2}.100\%=87,5\%\\ \%m_{H_2} = 100\% - 87,5\% = 12,5\%\)

a) PTHH: \(2CO+O_2\underrightarrow{t^o}2CO_2\)  (1)

                \(4H_2+O_2\underrightarrow{t^o}2H_2O\)  (2)

b) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{8,8}{44}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{O_2\left(1\right)}=0,1mol\\n_{O_2\left(2\right)}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CO}=0,1\cdot28=2,8\left(g\right)\\m_{H_2}=0,2\cdot2=0,4\left(g\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{CO}=\dfrac{2,8}{2,8+0,4}\cdot100\%=87,5\%\\\%m_{H_2}=12,5\%\end{matrix}\right.\)

c) PTHH: \(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo PTHH: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

\(\Rightarrow m_{KMnO_4}=0,6\cdot158=94,8\left(g\right)\)

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

PTHH : \(C+O_2\rightarrow CO_2\)

            x        x        x        (mol)

PTHH : \(S+O_2\rightarrow SO_2\)

           y          y          y       (mol)

\(\rightarrow\) x + y = 0,3 (1)

12x+32y=5,6  (2)

Từ (1),(2) \(\rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)

\(\%C=\dfrac{0,2.12}{5,6}.100=42,86\%\)

\(\%S=100\%-42,86\%=57,14\%\)

\(m_C=0,2.12=2,4\left(g\right)\)

\(m_S=5,6-2,4=3,2\left(g\right)\)

\(a)\\ C + O_2 \xrightarrow{t^o} CO_2\\ S + O_2 \xrightarrow{t^o} SO_2\\\)

\(b)\ n_C = a(mol) ; n_{S} = b(mol)\\ \Rightarrow 12a + 32b = 5,6 ; n_{O_2} = a + b = \dfrac{9,6}{32} = 0,3\\ \Rightarrow a = 0,2 ; b = 0,1\\ m_C = 0,2.12 = 2,4(gam) ; m_S = 0,1.32 = 3,2(gam)\\ c)\\ \Rightarrow \%m_C = \dfrac{0,2.12}{5,6}.100\% = 42,96\%\\ \%m_S = \dfrac{0,1.32}{5,6}.100\% = 57,14\%\)

\(d)\ n_{CO_2} = n_C = 0,2(mol)\\ n_{SO_2} = n_S = 0,1(mol)\\ \Rightarrow \%V_{CO_2} = \dfrac{0,2}{0,2 + 0,1}.100\% = 66,67\%\\ \%V_{SO_2} = 100\% - 66,67\% = 33,33\%\)

a/ \(2CO\left(0,2\right)+O_2\left(0,1\right)\rightarrow2CO_2\left(0,2\right)\)

\(2H_2\left(0,1\right)+O_2\left(0,05\right)\rightarrow2H_2O\left(0,1\right)\)

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

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

Số mol O₂ phản ứng ở phản ứng đầu là: \(0,15-0,05=0,1\)

\(\Rightarrow m_{CO_2}=0,2.44=8,8\)

b/ \(m_{CO}=0,2.28=5,6\)

\(m_{H_2}=0,1.2=0,2\)

c/ \(\%CO=\frac{0,2}{0,3}.100\%=66,67\%\)

\(\Rightarrow\%H_2=100\%-66,67\%=33,33\%\)