\(a.C_2H_5OH+3O_2-^{t^o}\rightarrow2CO_2+3H_2O\\ n_{C_2H_5OH}=0,3\left(mol\right)\\ n_{CO_2}=2n_{C_2H_5OH}=0,6\left(mol\right)\\ \Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\\ b.n_{O_2}=3n_{C_2H_5OH}=0,6\left(mol\right)\\ MàV_{O_2}=\dfrac{1}{5}V_{kk}\\ \Rightarrow V_{kk}=V_{O_2}.5=0,6.22,4.5=67,2\left(l\right)\\ c.n_{NaOH}=0,9\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,9}{0,6}=1,5\\ \Rightarrow Tạora2muốiNaHCO_3vàNa_2CO_3\\ Đặt:n_{NaHCO_3}=x\left(mol\right);n_{Na_2CO_3}=y\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}x+y=0,6\left(BTnguyento\left(C\right)\right)\\x+2y=0,9\left(BTnguyento\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,3\\y=0,3\end{matrix}\right.\\ \Rightarrow m_{muối}=0,3.84+0,3.106=57\left(g\right)\)

a)C2H6O+3O2to→2CO2+3H2O

0,3-------------0,9------0,6

nC2H6O=\(\dfrac{13,8}{46}\)=0,3mol

nCO2=2nC2H6O=0,6mol

⇒VCO2=0,6×22,4=13,44l

⇒Vkk=0,9×22,4×5=100,8l

n_Fe = 16.8/56 = 0.3 (mol) 

3Fe + 2O₂ -to-> Fe₃O₄ 

0.3......0.2...........0.1

VO2 = 0.2*22.4 = 4.48 (l) 

mFe3O4 = 0.1*232 = 23.2 (g) 

\(a) 2Cu + O_2 \xrightarrow{t^o} 2CuO\\ b)\\ n_{Cu} = n_{CuO} = \dfrac{8}{80} = 0,1(mol)\\ m_{Cu} = 0,1.64 = 6,4(gam)\\ c)\\ n_{O_2} = \dfrac{1}{2}n_{CuO} = 0,05(mol)\\ V_{O_2} = 0,05.22,4 = 1,12(lít)\\ d)\\ V_{không\ khí} = 5V_{O_2} = 1,12.5 = 5,6(lít)\)

Cảm ơn bạn nha

4P2+5)2 ---> 2P2O5

Lần lượt tính mol các chất theo tỉ lệ là đc

d) P2O5 + 3H2O ---> 2H3PO4

\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\ PTHH:4Al+3O_2\underrightarrow{t^o}2Al_2O_3\\ Mol:0,1\rightarrow0,075\rightarrow0,05\\ \rightarrow\left\{{}\begin{matrix}V_{kk}=\dfrac{0,075.22,4}{20\%}=8,4\left(l\right)\\m_{Al_2O_3}=0,05.102=5,1\left(g\right)\end{matrix}\right.\)

a.\(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2mol\)

\(2C_4H_{10}+13O_2\rightarrow\left(t^o\right)8CO_2+10H_2O\)

    0,2         1,3                                           ( mol )

\(V_{O_2}=1,3.22,4=29,12l\)

\(V_{kk}=29,12.5=145,6l\)

b.\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

        2,6                                                        1,3     ( mol )

\(m_{KMnO_4}=2,6.158=410,8g\)

\(n_{C_4H_{10}}=\dfrac{4,48}{22,4}=0,2\left(MOL\right)\\ pthh:2C_2H_{10}+13O_2\underrightarrow{t^o}8CO_2+10H_2O\) 
           0,2         1,3  
=> \(V_{O_2}=1,3.22,4=29,12\left(l\right)\\ V_{kk}=29,12:20\%=145,6\left(l\right)\\ pthh:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\) 
                  2,6                                              1,3 
=> \(m_{KMnO_4}=2,6.158=410,8\left(g\right)\)

  Sry chữ hơi xấu 🙂

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

\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ Mol:0,4\rightarrow\dfrac{4}{15}\rightarrow\dfrac{2}{15}\)

\(\rightarrow\left\{{}\begin{matrix}V_{O_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\rightarrow V_{kk}=\dfrac{448}{75}.5=\dfrac{448}{15}\left(l\right)\\m_{Fe_3O_4}=\dfrac{2}{15}.232=\dfrac{464}{15}\left(g\right)\end{matrix}\right.\)

2KClO₃ --to--> 2KCl + 3O₂

\(\dfrac{8}{45}\)                                \(\dfrac{4}{15}\)

\(m_{KClO_3}=\dfrac{8}{45}.122,5=\dfrac{196}{9}\left(g\right)\)

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

PTHH: 3Fe + 2O₂ ---t^o---> Fe₃O₄

Mol:     0,4      \(\dfrac{0,8}{3}\)               \(\dfrac{0,4}{3}\)

b, \(V_{O_2}=\dfrac{0,8}{3}.22,4=5,973\left(l\right)\)

c, \(V_{kk}=\dfrac{448}{75}.5=29,867\left(l\right)\)

d, \(m_{Fe_3O_4}=\dfrac{0,4}{3}.232=30,93\left(g\right)\)

e,

PTHH: 2KClO₃ ---t^o---> 2KCl + 3O₂

Mol:       \(\dfrac{0,16}{9}\)                            \(\dfrac{0,8}{3}\)

\(m_{KClO_3}=\dfrac{0,16}{9}.122,5=2,178\left(g\right)\)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)
=> pthh :\(3Fe+2O_2-t^o->Fe_3O_{\text{4 }}\)

             0,4       0,26              0,13 
=> \(V_{O_2}=0,26.22,4=5,973\left(L\right)\)
=> \(V_{KK}=5,973:\dfrac{1}{5}=29,86\left(L\right)\)
\(m_{Fe_3O_4}=0,13.232=30,16\left(g\right)\)
\(pthh:2KClO_3-t^o->2KCl+3O_2\)
           0,173                                0,26 
=> \(m_{KClO_3}=0,173.122,5=21,1925\left(G\right)\)

a, PT: \(2Zn+O_2\underrightarrow{t^o}2ZnO\) - pư hóa hợp.

b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\), ta được Zn dư.

Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(n_{ZnSO_4}=n_{H_2SO_4}=0,1\left(mol\right)\Rightarrow m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)