Đốt cháy hoàn toàn 28 lít hỗn hợp gồm CH4 và C2H2 trong đó CH4 chiếm 20% về thể tích. Hãy tính
a) Thể tích không khí cần dùng biết oxi chiếm 20% thể tích không khí
b) Thể tích khí CO2 tạo thành biết các khí đều đo ở đktc
-----------------------------
CH4 chiếm 20% trong 28 lit hỗn hợp
=> V CH4 = 20%.28 = 5,6 lit
=> nCH4 = V/22,4 = 0,25 mol
=> V C2H2 = 28 - 5,6 = 22,4 lit
=> n C2H2 = V/22,4 = 1 mol
CH4       +   2O2 ---------> CO2 + 2H2O 
0,25           0,5                0,25
C2H2     +    5/2O2 -------------> 2CO2 + H2O
1                2,5                            2
=> V kh + 2,5).100/20.22,4 = 336 lit
=> VCO2 = (0,25 + 2).22,4 = 50,4 lit

\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

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

0,3      0,2       0,1

\(V_{O_2}=0,2\cdot22,4=4,48l\)

\(V_{kk}=5V_{O_2}=5\cdot4,48=22,4l\)

\(m_{O_2}=0,2\cdot32=6,4g\)

\(m_{Al}=27,8.19,2\%=5,4\left(g\right)\\ m_{Fe}=27,8-5,4=22,4\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\end{matrix}\right.\)

PTHH:

4Al + 3O₂ --t^o--> 2Al₂O₃

0,2-->0,15------->0,1

3Fe + 2O₂ --t^o--> Fe₃O₄

0,4-->4/15--------->2/15

\(\rightarrow\left\{{}\begin{matrix}V_{kk}=\left(0,15+\dfrac{4}{15}\right).22,4.5=\dfrac{140}{3}\left(l\right)\\m_{Cran}=0,1.102+\dfrac{2}{15}.232=\dfrac{617}{15}\left(g\right)\end{matrix}\right.\)

mAl=27,8.19,42%=5,4g

⇒nAl=\(\dfrac{5,4}{27}\)=0,2mol

⇒nFe=\(\dfrac{27,8-5,4}{56}\)=0,4mol

4Al+3O2to→2Al2O34

3Fe+2O2to→Fe3O4

⇒nO2=\(\dfrac{3}{4}\)nAl+\(\dfrac{2}{3}\)nFe=\(\dfrac{5}{12}\)mol

⇒Vkk=\(\dfrac{5}{12}\).22,4.5=46,67l

b,

mrắn=27,8+mO2=27,8+​​\(\dfrac{5}{12}\)32=41,1g

→⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩Vkk=(0,15+415

\(m_C=10.36\%=3,6\left(g\right)\Rightarrow n_C=0,3\left(mol\right)\)

\(m_S=10-3,6=6,4\left(g\right)\Rightarrow n_S=0,2\left(mol\right)\)

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

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

\(n_{O_2}=0,3+0,2=0,5\left(mol\right)\)

Vì oxi chiếm 20% thể tích không khí 

=> \(V_{kk}=\dfrac{0,5.22,4}{20\%}=56\left(lít\right)\)

\(n_{hh}=n_{CO_2}+n_{SO_2}=0,3+0,2=0,5\left(mol\right)\)

=> \(V_{hh}=0,5.22,4=11,2\left(lít\right)\)

PTHH: 

mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)mS=10−3,6=6,4(g)⇔nS=6,432=0,2(mol)

VO2=(0,3+0,2)22,4=11,2(l)VO2=(0,3+0,2)22,4=11,2(l)

mhh=mCO2+mSO2=0,3.44+0,2.64=26(g)

\(m_{Al}=19,2\%.27,8=5,3376\left(g\right)\Rightarrow n_{Al}=0,2\left(mol\right)\)

\(m_{Fe}=27,8-5,3376=22,4624\left(g\right)\Rightarrow n_{Fe}=0,4\left(mol\right)\)

\(4Al+3O_2-^{t^o}\rightarrow2Al_2O_3\)

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

Theo PT : \(n_{O_2}=0,2.\dfrac{3}{4}+0,4.\dfrac{2}{3}=\dfrac{5}{12}\left(mol\right)\)

Vì oxi chiếm 20% thể tích không khí

=> \(V_{kk}=\dfrac{5}{12}.22,4.\dfrac{100}{20}=\dfrac{140}{3}\left(lít\right)=46,67\left(lít\right)\)

Bảo toàn khối lượng ta có: \(m_{KL}+m_{O_2}=m_{oxit}\)

=> \(m_{oxit}=27,8+\dfrac{5}{12}.32=\dfrac{617}{15}\left(g\right)=41,13\left(g\right)\)

a, 

mAl=27,8.19,42%=5,4gmAl=27,8.19,42%=5,4g

⇒nAl=5,427=0,2mol⇒nAl=5,427=0,2mol

⇒nFe=27,8−5,456=0,4mol⇒nFe=27,8−5,456=0,4mol

4Al+3O2to→2Al2O34Al+3O2→to2Al2O3

3Fe+2O2to→Fe3O43Fe+2O2→toFe3O4

⇒nO2=34nAl+23nFe=512mol⇒nO2=34nAl+23nFe=512mol

⇒Vkk=512.22,4.5=46,67l⇒Vkk=512.22,4.5=46,67l

b,

mrắn=27,8+mO2=27,8+512.32=41,1g

\(m_{Al}=27,8.\dfrac{19,2}{100}=5,4\) ( g )

\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\)  ( mol ) 

\(m_{Fe}=27,8-5,4=22,4\) ( g )

\(n_{Fe}=\dfrac{m}{m}=\dfrac{22,4}{56}=0,4\) ( mol )

\(4Al\) + \(3O_2\) →  \(2Al_2O_3\)

\(3Fe\) + \(2O_2\) →  \(Fe_3O_4\)

\(n_{O_2}=\dfrac{3}{4}Al+\dfrac{2}{3}Fe=\dfrac{3}{4}.0,2+\dfrac{2}{3}.0,4=\dfrac{5}{12}\)  ( mol )

\(V_{kk}=5V_{O_2}=5.\dfrac{5}{12}.22,4\approx46,7\) ( \(l\) )

0,2 mol                    →       0,1 mol

\(a,n_{Al}=\dfrac{19,2\%.27,8}{27}=\dfrac{1112}{5625}\left(mol\right)\\ n_{Fe}=\dfrac{\left(100\%-19,2\%\right).27,8}{56}=\dfrac{14039}{35000}\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{O_2\left(tổng\right)}=\dfrac{1112}{5625}.0,75+\dfrac{2}{3}.\dfrac{14039}{35000}\approx0,4156762\left(mol\right)\\ V_{kk\left(đktc\right)}\approx0,4156762.5.22,4\approx46,5557344\left(l\right)\\ b,n_{Al_2O_3}=\dfrac{\dfrac{1112}{5625}}{2}=\dfrac{556}{5625}\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{\dfrac{14039}{35000}}{3}=\dfrac{14039}{105000}\left(mol\right)\\ m_{rắn}=\dfrac{14039}{105000}.232+\dfrac{556}{5625}.102=41,1016381\left(g\right)\)

Fe hóa trị mấy để tạo ra h/c

\(m_{Fe}=32.70\%=22,4\left(g\right)\\ n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ m_{Mg}=32-22,4=9,6\left(g\right)\\ n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ 2Mg+O_2\rightarrow\left(t^o\right)2MgO\\ n_{O_2\left(tổng\right)}=\dfrac{2}{3}.n_{Fe}+\dfrac{1}{2}.n_{Mg}=\dfrac{2}{3}.0,4+\dfrac{1}{2}.0,4=\dfrac{7}{15}\left(mol\right)\\ V_{O_2\left(đktc\right)}=\dfrac{7}{15}.22,4=\dfrac{784}{75}\left(l\right)\\ V_{kk\left(đktc\right)}=\dfrac{784}{75}.\dfrac{100}{20}=\dfrac{784}{15}\left(l\right)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}.0,4=\dfrac{2}{15}\left(mol\right)\\ n_{MgO}=n_{Mg}=0,4\left(mol\right)\\ m_{rắn}=m_{Fe_3O_4}+m_{MgO}=\dfrac{232.2}{15}+0,4.40=\dfrac{704}{15}\left(g\right)\)

\(a,m_{Fe}=\%Fe.m_{hh}=70\%.32=22,4\left(g\right)\)

\(m_{Mg}=m_{hh}-m_{Fe}=32-22,4=9,6\left(g\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{22,4}{56}=0,4\left(mol\right)\\ n_{Mg}=\dfrac{m}{M}=0,4\left(mol\right)\)

\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\left(1\right)\)

\(PTHH:2Mg+O_2\underrightarrow{t^o}2MgO\left(2\right)\)

\(Theo.PTHH\left(1\right):n_{O_2\left(Fe\right)}=\dfrac{2}{3}n_{Fe}=\dfrac{2}{3}.0,4=\dfrac{4}{15}\left(mol\right)\)

\(Theo.PTHH\left(2\right):n_{O_2\left(Mg\right)}=2n_{Mg}=2.0,4=0,8\left(mol\right)\)

\(\Rightarrow n_{O_2\left(tổng\right)}=n_{O_2\left(Fe\right)}+n_{O_2\left(Mg\right)}=\dfrac{4}{15}+0,8=\dfrac{16}{15}\left(mol\right)\)

\(\Rightarrow V_{O_2\left(đktc\right)}=n.22,4=\dfrac{16}{15}.22,4=\dfrac{1792}{75}\left(l\right)\)

\(\Rightarrow V_{kk}=V_{O_2}.5=\dfrac{1792}{75}.5=\dfrac{1792}{15}\left(l\right)\)

PTHH: \(2K+2H_2O\rightarrow2KOH+H_2\)

\(n_K=\dfrac{7,8}{39}=0,2\left(mol\right)\)

Theo PTHH: \(n_{H_2}=\dfrac{1}{2}n_K=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

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

Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

\(\Rightarrow V_{O_2}=n_{O_2}.22.4=0,05.22,4=1,12\left(l\right)\)

Thể tích không khí đã dùng:

\(V_{kk}=\dfrac{V_{O_2}.100\%}{20\%}==5,6\left(l\right)\)

ko phải e làm đâu nha a , a tham khảo cho 

a)CH4 + 2O2 →→ CO2 + 2H2O (1)

2H2 + O2 →→ 2H2O (2)

b) Đặt nCH4 = a (mol) , nH2 = b (mol)

=> mCH4 = 16a(g) , mH2 = 2b(g)

mà n(CH4 + H2) = V/22,4 = 6,72/22,4 = 0,3(mol)

=> a + b = 0,3(mol) => a = 0,3 - b

Từ PT(1) => nO2 = 2nCH4 = 2a(mol)

Từ PT(2) => nO2 = 1/2 . nH2 = 1/2 . b (mol)

=>tổng nO2 = 2a + 1/2.b (mol)

=> mO2 = n .M = 32. (2a+ 1/2.b )= 64a + 16b(g)

Theo ĐLBTKL:mCH4 + mH2 + mO2 = mCO2 + mH2O(PT1,2)

=> 16a + 2b+ 64a +16b = 11,6

=> 80a + 18b = 11,6

=> 80. (0,3 - b ) + 18b = 11,6

=> b = 0,2(mol)

=> a = 0,3 - 0,2 = 0,1(mol)

=>mCH4 = 16a = 16. 0,1 = 1,6(g) , mH2 = 2b = 2 . 0,2 = 0,4(g)

=> VCH4 = n .22,4 = 0,1 . 22,4 =2,24(l)

VH2 = n . 22,4 = 0,2 . 22,4 = 4,48(l)

tự tính ... 

c) nO2 = 2a + 1/2 .b = 2 . 0,1 + 1/2 . 0,2 = 0,3(mol)

=> VO2 = 0,3 . 22,4 =6,72(l)

mà VO2 = 20% Vkk

=> Vkk = 6,72 : 20% =33,6(l)