Bài 1:

a) 2CuFeS₂ + \(\dfrac{13}{2}\)O₂ --t^o--> 2CuO + Fe₂O₃ + 4SO₂

b) \(n_{CuFeS_2}=\dfrac{3,68}{184}=0,02\left(mol\right)\)

\(n_{O_2}=\dfrac{1,68}{22,4}=0,075\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,02}{2}< \dfrac{0,075}{\dfrac{13}{2}}\) => CuFeS₂ hết, O₂ dư

PTHH: 2CuFeS₂ + \(\dfrac{13}{2}\)O₂ --t^o--> 2CuO + Fe₂O₃ + 4SO₂

              0,02----->0,065------->0,02---->0,01---->0,04

=> \(\left\{{}\begin{matrix}n_{O_2\left(dư\right)}=0,075-0,065=0,01\left(mol\right)\\n_{SO_2}=0,04\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,01}{0,01+0,04}.100\%=20\%\\\%V_{SO_2}=\dfrac{0,04}{0,01+0,04}.100\%=80\%\end{matrix}\right.\)

- \(\left\{{}\begin{matrix}m_{CuO}=0,02.80=1,6\left(g\right)\\m_{Fe_2O_3}=0,01.160=1,6\left(g\right)\end{matrix}\right.\)

=> m_rắn = 1,6 + 1,6 = 3,2 (g)

Bài 2: 

a) 

2CuS + 3O₂ --t^o--> 2CuO + 2SO₂

4FeS + 7O₂ --t^o--> 2Fe₂O₃ + 4SO₂

b) Gọi số mol CuS, FeS là a, b (mol)

=> 96a + 88b = 22,8 (1)

\(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

=> a + b = 0,25 (2)

(1)(2) => a = 0,1; b = 0,15

=> \(\left\{{}\begin{matrix}n_{CuO}=0,1\left(mol\right)\\n_{Fe_2O_3}=0,075\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{CuO}=\dfrac{0,1.80}{0,1.80+0,075.160}.100\%=40\%\\\%m_{Fe_2O_3}=\dfrac{0,075.160}{0,1.80+0,075.160}.100\%=60\%\end{matrix}\right.\)

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

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

\(n_{CH_4}=\dfrac{6,4}{16}=0,4mol\)

\(CH_4+2O_2\rightarrow\left(t^o\right)CO_2+2H_2O\)

0,4        0,8               0,4                   ( mol )

\(m_{CO_2}=0,4.44=17,6g\)

\(V_{O_2}=0,8.22,4=17,92l\)

bạn từng câu lên sẽ dễ nhìn hơn 

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

Ta có: \(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\)

a, Theo PT: \(n_{O_2}=\dfrac{5}{4}n_P=0,125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,125.22,4=2,8\left(l\right)\)

b, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,05\left(mol\right)\)

\(\Rightarrow m_{P_2O_5}=0,05.142=7,1\left(g\right)\)

Bạn tham khảo nhé!

\(n_P=\dfrac{3.1}{31}=0.1\left(mol\right)\)

\(4P+5O_2\underrightarrow{t^0}2P_2O_5\)

\(0.1...0.125....0.05\)

\(V_{O_2}=0.125\cdot22.4=2.8\left(l\right)\)

\(m_{P_2O_5}=0.05\cdot142=7.1\left(g\right)\)

PTHH : 4P + 5\(O_2\) -> 2\(P_2O_5\)

a) nP = m / M = 0.1(mol)

có : nP = 5/4 n\(O_2\)

    -> n\(O_2\) = 0.125 (mol)

   -> V\(O_2\) (đktc) = n.22.4 = 2.8 (L)

b) có : nP = 1/2 n\(P_2O_5\)

   -> n\(P_2O_5\) = 0.05 (mol)

   -> m\(P_2O_5\) = n.M = 7.1 (g)

nCH4 = 3,36 : 22,4 = 0,15 (mol) 
pthh : CH4 + 2O2 -t--> CO2 + 2H2O 
         0,15      0,3          
=> VO2 = 0,3 . 22,4 = 6,72 (L) 
ta có  : VO2 = 20% Vkk => Vkk = VO2 : 20% = 6,72 : 20% = 33,6 (L)

\(n_{CH_4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\\ Mol:0,15\rightarrow0,3\\ \rightarrow\left\{{}\begin{matrix}V_{O_2}=0,3.22,4=6,72\left(l\right)\\V_{kk}=6,72.5=33,6\left(l\right)\end{matrix}\right.\)

\(n_P=\dfrac{7,44}{31}=0,24mol\)

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

0,24   0,3      0,12

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

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)

0,2                           0,3

\(m_{KClO_3}=0,2\cdot122,5=24,5g\)

a. \(n_P=\dfrac{7.44}{31}=0,24\left(mol\right)\)

PTHH : 4P + 5O₂ ------t^o----> 2P₂O₅

            0,24    0,3

b. \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c. 2KClO₃ ---t^o---> 2KCl + 3O₂

       0,2                              0,3

\(m_{KClO_3}=0,2.122,5=24,5\left(g\right)\)