Ta có: \(m_C=1,5.1000.90\%=1350\left(g\right)\)

\(n_C=\dfrac{1350}{12}=112,5\left(mol\right)\)

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

Theo PT: \(n_{O_2}=n_C=112,5\left(mol\right)\)

\(\Rightarrow V_{O_2}=112,5.22,4=2520\left(l\right)\)

\(V_{kk}=V_{O_2}.5=12600\left(l\right)\)

B1:

\(n_C=\dfrac{96\%.14.1000}{12}=1120\left(mol\right)\\ n_S=\dfrac{2,56\%.14.1000}{32}=11,2\left(mol\right)\\ C+O_2\rightarrow\left(t^o\right)CO_2\\ S+O_2\rightarrow\left(t^o\right)SO_2\\ n_{SO_2}=n_S=11,2\left(mol\right)\\ n_{CO_2}=n_C=1120\left(mol\right)\\ V_{CO_2\left(đktc\right)}=1120.22,4=25088\left(l\right)\\ n_{SO_2\left(đktc\right)}=11,2.22,4=250,88\left(l\right)\)

B2:

\(n_{C_2H_6}=\dfrac{1,8.\left(100\%-2\%\right).1000}{22,4}=78,75\left(mol\right)\\ C_2H_6+\dfrac{7}{2}O_2\rightarrow\left(t^o\right)2CO_2+3H_2O\\ n_{O_2\left(đktc\right)}=\dfrac{7}{2}.78,75=275,625\left(mol\right)\\ V_{O_2\left(đktc\right)}=275,625.22,4=6174\left(l\right)=6,174\left(m^3\right)\)

→⎧⎪ ⎪⎨⎪ ⎪⎩nC=60012=50(mol)nS=832=0,25(mol)→{nC=60012=50(mol)nS=832=0,25(mol)

PTHH:

C + O₂ ---t^o----> CO₂

_{50.....50.......................................(mol)}

S + O₂ ----t^o---> SO₂

^{0,25...0,25 ............................................(mol)}

→ VO2VO2 _{(cần dùng)} = (50 + 0,25) . 22,4 = 1125,6 (l)

=>Vkk=1125.6.5=5628l

\(n_{P_2O_5}=\dfrac{21,3}{142}=0,15\left(mol\right)\)

PTHH: 4P + 5O₂ --to--> 2P₂O₅

            0,3    0,375         0,15

\(\rightarrow\left\{{}\begin{matrix}m_P=0,3.31=9,3\left(g\right)\\V_{O_2}=0,375.22,4=8,4\left(l\right)\\V_{kk}=8,4.5=42\left(l\right)\end{matrix}\right.\)

PTHH: 2KClO₃ --to--> 2KCl + 3O₂

          0,25                              0,375

=> m_KClO3 = 0,25.122,5 = 30,625 (g)

\(nP_2O_5=\dfrac{21,3}{142}=0,15\left(mol\right)\)

\(pthh:4P+5O_2-t^o->2P_2O_5\)
           0,3     0,375                0,15
=> \(m_P=0,3.31=9,3\left(g\right)\)
=>\(V_{O_2}=0,375.22,4=8,4\left(L\right)=>V_{KK}=8,4:20\%=42\left(L\right)\)
\(pthh:2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\) 
           0,75                                                 0,75 
=> mKMnO4 = 0,75 . 158 = 118,5 (G)                                                   

Khối lượng C chứa trong 1 tấn than:

PTHH: \(C+O_2\xrightarrow[]{t^o}CO_2\)

           12---32-------gam

           0,96--x-------tấn

\(\Rightarrow x=\dfrac{0,96.32}{12}=2,56\left(\text{tấn}\right)\)

C₂H₂+\(\dfrac{5}{2}\)O₂-to>2CO₂+H₂O

0,05--0,125---------0,1 mol

n _C2H2=\(\dfrac{1,12}{22,4}\)=0,05 mol

=>m _CO2=0,1.44=4,4g

=>Vkk=0,125.5.22,4=5,6l

CH₄+2O₂-to>CO2+2H₂O

0,042-0,084 mol

=>CH4 thực tế =0,95 m³

=>n CH4=\(\dfrac{0,95}{22,4}\)=0,042 mol

=>VO2=0,084.22,4=1,8816 (m³)

=>Vkk=1,8816.5=9,408(m³)

\(n_{CO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(C_2H_5OH+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

0,2               0,6       0,4         0,6

a)\(m_{C_2H_5OH}=0,2\cdot46=9,2g\)

b)\(V_{O_2}=0,6\cdot22,4=13,44l\)

\(\Rightarrow V_{kk}=5V_{O_2}=5\cdot13,44=67,2l\)

\(n_{CO_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH : \(C+O_2\underrightarrow{t^0}CO_2\)

PT :      1mol  1mol

Đề :      0,4mol ?mol

=> \(n_{O_2}=\frac{0,4\cdot1}{1}=0,4\left(mol\right)\)

=> \(V_{O_2}=0,4\cdot22,4=8,96\left(l\right)\)

\(V_{kk}\cdot20\%=V_{O_2}\Rightarrow V_{kk}=\frac{V_{O_2}}{20\%}=\frac{8,96}{20\%}=44,8\left(l\right)\)

=> \(V_{kk}=44,8l\)