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

\(n_P=\dfrac{6,2}{31}=0,2mol\\n_{O_2}=\dfrac{0,2.5}{4}=0,25mol \)

\(S+O_2\underrightarrow{t^o}SO_2\)

\(n_S=\dfrac{3,2}{32}=0,1mol\\ n_{O_2}=0,1mol\)

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

\(n_C=\dfrac{2,4}{12}=0,2mol\\ n_{O_2}=0,2mol\\ n_{O_2}\left(tổng\right)=\)

\(0,25+0,1+0,2=0,55mol\\ m_{O_2}\left(trong.hh.B\right)=0,55.32=17,6g\)

a, \(m_{Fe}=0,25.56=14g\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(\Rightarrow n_{O_2}=\dfrac{0,25.2}{3}=0,16mol\\ m_{O_2}=0,16.32=5,12g\)

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

\(n_{O_2}=\dfrac{0,25.3}{4}=0,1875mol\\ m_{O_2}=0,1875.32=6g\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

\(n_{O_2}=\dfrac{0,5.1}{2}=0,25mol\\ m_{O_2}=0,25.32=8g\)

\(\Rightarrow m_{O_2}\left(trong.hỗn.hợp.A\right)=\) \(5,12+6+8=19,12g\)

help mi

nH2 = 8 : 2 = 4 (mol) 
nCH4 = 2 : 16 = 0,125 (mol) 
pthh : 2H2 + O2-t--> 2H2O 
         4------->2 (mol) 
       CH4 + 2O2 ----> CO2 + 2H2O 
        0,125->0,25 (MOL) 
=> nO2= 2 + 0,25 = 2,25 (mol) 
   => mO2 = 2,25 . 32 = 72 (g) 

C  +  O2 --t--> CO2 
0,15->0,15 (mol) 
S  +   O2 -t--> SO2 
0,125->0,125 (mol) 
=> nO2 = 0,15 + 0,125 = 0,275 (mol) 
=> mO2 = 0,275 . 32 = 8,8 (G)

a) 3Fe+2O2--->FE3O4

0,5------1/3 (mol)

4Al+3O2---.2Al2O3

1,25--0,9375(mol)

2Zn+O2--->2ZnO

1,5---0,75(mol)

n O2=1/3+0,9375+0,75=2,02(mol)

m O2=2,02.32=64,64(g)

b) 4P+5O2-->2P2O5

0,1-----0,125(mol)

S+02--->SO2

0,2--0,2(mol)

C+O2-->CO2

0,3--0,3(mol)

n O2=0,125+0,2+0,3=0,625(mol)

m O2=0,625.32=20(g)

c) n CH4=1,6/16=0,1(mol)

n CO=2,8/28=0,1(mol)

n C4H10=0,58/58=0,01(mol)

CH4+2O2--->CO2+2H2O

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

2CO+O2-->2CO2

0,1--0,05(mol)

C4H10+13/2O2--->4CO2+5H2O

0,01-----0,065(mol)

n O2=0,2+0,05+0,065=0,315(mol)

m O2=0,315.32=10,08(g)

a. \(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\)

PTHH : 3Fe + 2O₂ -t^o-> Fe₃O₄

              0,09    0,06        0,03

\(m_{Fe}=0,09.56=5,04\left(g\right)\)

\(V_{O_2}=0,06.22,4=1,344\left(l\right)\)

b. PTHH : 2KCl + 3O₂ -> 2KClO₃

                            0,06           0,04

\(m_{KClO_3}=0,04.122,5=4,9\left(g\right)\)

4

n _Fe3O4=\(\dfrac{6,96}{232}=0,03mol\)

3Fe+2O₂-to>Fe₃O₄

0,09---0,06-----0,03 mol

=>m _Fe=0,09.56=5,04g

=>V_O2=0,06.22,4=1,344l

b) 

2KClO₃-to>2KCl+3O₂

0,04----------------------0,06 mol

=>m _KClO3=0,04.122,5=4,9g

\(n_{Fe_3O_4}=\dfrac{6,96}{232}=0,03\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=3.0,03=0,09\left(mol\right);n_{O_2}=0,03.2=0,06\left(mol\right)\\ a,\Rightarrow m_{Fe}=0,09.56=5,04\left(g\right);V_{O_2\left(đktc\right)}=0,06.22,4=1,344\left(l\right)\\ 2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\uparrow\\ n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2.0,06}{3}=0,04\left(mol\right)\\ \Rightarrow b,m_{KClO_3}=122,5.0,04=4,9\left(g\right)\)

\(2H_2 + O_2 \xrightarrow{t^o} 2H_2O\\ n_{O_2} = \dfrac{1}{2}n_{H_2} = 2,5(mol)\\ m_{O_2} = 2,5.32 = 80(gam)\)

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

\(5......2.5\)

\(m_{O_2}=2.5\cdot32=80\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\x_{Ca}=y\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}24x+40y=17,6\\x=2y\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,4\\y=0,2\end{matrix}\right.\)

a)\(m_{Mg}=0,4\cdot24=9,6g\)

   \(m_{Ca}=0,2\cdot40=8g\)

b)\(2Mg+O_2\underrightarrow{t^o}2MgO\)

   \(2Ca+O_2\underrightarrow{t^o}2CaO\)

Từ hai pt: \(\Rightarrow\Sigma n_{O_2}=\dfrac{1}{2}n_{Mg}+\dfrac{1}{2}n_{Ca}=\dfrac{1}{2}\cdot0,4+\dfrac{1}{2}\cdot0,2=0,3mol\)

\(\Rightarrow m_{O_2}=0,3\cdot32=9,6g\)

\(V_{O_2}=0,3\cdot22,4=6,72l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot6,72=33,6l\)

a) 

Có \(\left\{{}\begin{matrix}24.n_{Mg}+40.n_{Ca}=17,6\\n_{Mg}=2.n_{Ca}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{Ca}=0,2\left(mol\right)\\n_{Mg}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Ca}=0,2.40=8\left(g\right)\\m_{Mg}=0,4.24=9,6\left(g\right)\end{matrix}\right.\)

b)

PTHH: 2Ca + O₂ --t^o--> 2CaO 

            0,2-->0,1

             2Mg + O₂ --t^o--> 2MgO

            0,4--->0,2

=> \(V_{O_2}=\left(0,1+0,2\right).22,4=6,72\left(l\right)\)

\(V_{kk}=6,72.5=33,6\left(l\right)\)

\(n_{Fe}=\dfrac{1,68}{56}=0,03\left(mol\right)\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

 0,03   0,02               0,01  ( mol )

\(m_{Fe_3O_4}=0,01.232=2,32\left(g\right)\)

\(V_{kk}=0,02.22,4.5=2,24\left(l\right)\)

\(2KClO_3\rightarrow\left(t^o\right)2KCl+3O_2\)

   \(\dfrac{1}{75}\)                               0,02   ( mol )

\(m_{KClO_3}=\dfrac{1}{75}.122,5=1,63\left(g\right)\)

a,nFe=1,68/56=0,03 mol

Ta có PTHH : 3Fe + 2O2 --> Fe3O4 (1) ( ở trên dấu --> có to nha )

Theo PTHH ta có :

nFe3O4=1/3nFe=1/3.0,03=0,01 mol

nO2=2/3nFe=2/3.0,03=0,02 mol

=>mFe3O4= 0,01.232=2,32g

=>Vkk=5.(0,02.22,4)=2,24 l

b, Ta có PTHH: 2KClO3 --> 2KCl + 3O2 (2) ( trên dấu --> vẫn có to )

Gọi x là số mol KClO3 cần dùng ( x > 0 )

Theo PTHH (3) và theo bài ra ta có PTHH sau:

2/3x=0,02 

=> x=0,03 mol

=> mKClO3= 0,03.122,5= 3,675g