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

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 + O2 -> CO2
a......a........a
S + O2 -> SO2
b.......b.........b
=> 12a + 32b = 5,6
(a + b )32 = 9,6
giải ra: a = 0,2 ; b = 0,1
m C = 0,2 . 12 = 2,4g
m S = 0,1 .32 = 3,2 g
b. 

% n SO2 = 100% - 66,67% = 33,33%

a) \(n_{CH_4}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

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

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

          0,05-->0,1------->0,05

             2C₂H₂ + 5O₂ --t^o--> 4CO₂ + 2H₂O

            0,125<--0,3125<----0,25

=> \(\left\{{}\begin{matrix}\%V_{CH_4}=\dfrac{0,05}{0,05+0,125}.100\%=28,57\%\\\%V_{C_2H_2}=\dfrac{0,125}{0,05+0,125}.100\%=71,43\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\%m_{CH_4}=\dfrac{0,05.16}{0,05.16+0,125.26}.100\%=19,753\%\\\%m_{C_2H_2}=\dfrac{0,125.26}{0,05.16+0,125.26}.100\%=80,247\%\end{matrix}\right.\)

b) \(n_{O_2}=0,1+0,3125=0,4125\left(mol\right)\)

=> \(V_{O_2}=0,4125.22,4=9,24\left(l\right)\)

=> V_kk = 9,24.5 = 46,2 (l)

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

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

_____0,2<---0,6<--------------0,3

=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{15,6}.100\%=34,615\%\\\%Al_2O_3=\dfrac{15,6-0,2.27}{15,6}.100\%=65,385\%\end{matrix}\right.\)

b) \(n_{Al_2O_3}=\dfrac{15,6-0,2.27}{102}=0,1\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl --> 2AlCl₃ + 3H₂O

______0,1--->0,6

=> n_HCl = 0,6+0,6 = 1,2(mol)

=> \(V_{dd}=\dfrac{1,2}{2}=0,6\left(l\right)\)

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%

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: \(n_{KMnO_4}=2n_{O_2}=0,6mol\)

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

theo bài ta có:nH2=6,72/22,4=0,3(mol)

pthh; 2Al+3H2SO4------>Al2(SO4)3+3H2 (1)

2Al2O3+3H2SO4-------->Al2(SO4)3+3H2O (2)

ta có:nAl=0,3(mol)

=>mAl=0,3*27=8,1(g)

=>mAl2O3=15,6-8,1=7,5(g)

=>nAl2O3=7,5/102=5/68(mol)

=>nH2SO4=1,5*0,3+1,5*5/68=0,56(mol)

=>V(H2SO4)=0,56/1,5=0,37(l)

Chúc bạn học tốt

a) \(n_{SO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

         0,5<-0,5<------0,5

=> m_S = 0,5.32 = 16(g)

=> \(\left\{{}\begin{matrix}\%m_S=\dfrac{16}{22,2}.100\%=72,07\%\\\%m_P=\dfrac{22,2-16}{22,2}.100\%=27,93\%\end{matrix}\right.\)

b) \(n_P=\dfrac{22,2-16}{31}=0,2\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

          0,2-->0,25----->0,1

=> \(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

c) 

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

             0,5<-------------------0,75

=> \(m_{KClO_3}=0,5.122,5=61,25\left(g\right)\)

a) PTHH:

\(S+O_2\rightarrow\left(t^o\right)SO_2\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\)

- Chất khí mùi hắc là SO₂

- Chất rắn sau phản ứng có m(g) là P₂O₅

Đặt: n_S=a(mol); n_P=b(mol) (a,b>0) (nguyên, dương)

\(\Rightarrow\left\{{}\begin{matrix}32a+31b=22,2\\22,4a=11,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\\b=0,2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_S=\dfrac{0,5.32}{22,2}.100\approx72,072\%\\\%m_P\approx100\%-72,072\%\approx27,928\%\end{matrix}\right.\)

b)

\(n_{O_2}=a+\dfrac{5}{4}b=0,5+\dfrac{5}{4}.0,2=0,75\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,75.22,4=16,8\left(l\right)\)

c)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,75}{3}=0,5\left(mol\right)\\ \Rightarrow m_{KClO_3}=122,5.0,5=61,25\left(g\right)\)