Bài 24: Tính chất của oxi

\(n_P=\dfrac{12,4}{31}=0,4\left(mol\right)\\ n_{O_2}=\dfrac{17}{32}=0,53125\left(mol\right)\\ 4P+5O_2\rightarrow\left(t^o\right)2P_2O_5\\ Vì:\dfrac{0,4}{4}< \dfrac{0,53125}{5}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=\dfrac{5}{4}.0,4=0,5\left(mol\right)\\ \Rightarrow m_{O_2\left(dư\right)}=32.\left(0,53125-0,5\right)=1\left(g\right)\\ n_{P_2O_5}=\dfrac{2}{4}.0,4=0,2\left(mol\right)\\ \Rightarrow m_{P_2O_5}=142.0,2=28,4\left(g\right)\)

\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

Pt : \(2Cu+O_2\underrightarrow{t^o}2CuO|\)

         2        1           2

        0,1     0,05       0,1

a) \(n_{O2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)

\(V_{O2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)

b) \(n_{CuO}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)

⇒ \(m_{CuO}=0,1.80=8\left(g\right)\)

 Chúc bạn học tốt

\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^o}CuO\)

\(n_{Cu}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1mol\)

\(V_{O_2}=n.22,4=0,1.\dfrac{1}{2}.22,4=1,12l\)

\(m_{sp}=n.M=0,1.80=8g\)

C₂H₄+3O₂-to>2CO₂+2H₂O

              0,3-------0,2-------0,2

n _C2H4=\(\dfrac{4,48}{22,4}\)=0,2 mol

n _O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>C2H4 dư

=>V_CO2=V_H2O=0,2.22,4=4,48l

=>V_{C2H4 dư}=0,1.22,4=2,24l

\(n_{O_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ C_2H_4+3O_2\rightarrow\left(t^o\right)2CO_2+2H_2O\\ Vì:\dfrac{0,2}{1}>\dfrac{0,3}{3}\Rightarrow C_2H_4dư\\ \Rightarrow n_{CO_2}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ n_{C_2H_4\left(dư\right)}=0,2-\dfrac{0,3}{3}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ V_{C_2H_4\left(dư\right)}=0,1.22,4=2,24\left(l\right)\)

Bài 6:

\(a,n_{Al}=\dfrac{6,75}{27}=0,25\left(mol\right)\\ n_{O_2}=\dfrac{8,4}{32}=0,2625\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,25}{4}< \dfrac{0,2625}{3}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=0,2625-\dfrac{3}{4}.0,25=0,075\left(mol\right)\\ b,n_{Al_2O_3}=\dfrac{2}{4}.n_{Al}=\dfrac{2}{4}.0,25=0,125\left(mol\right)\\ \Rightarrow m_{Al_2O_3}=102.0,125=12,75\left(g\right)\)

Bài 7:

\(a,n_{Fe}=\dfrac{7}{56}=0,125\left(mol\right)\\ n_{O_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ 3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ Vì:\dfrac{0,125}{3}< \dfrac{0,3}{2}\Rightarrow O_2dư\\ b,n_{Fe_3O_4}=\dfrac{1}{3}.n_{Fe}=\dfrac{0,125}{3}=\dfrac{1}{24}\left(mol\right)\\ \Rightarrow m_{sp}=m_{Fe_3O_4}=\dfrac{232}{24}=\dfrac{29}{3}\left(g\right)\)

a.\(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

PTHH : 4P + 5O₂ \(\underrightarrow{t^0}\)2P₂O₅

            0,2      0,25

\(V_{O_2}=0,25.22,4=5,6\left(l\right)\)

4P+5O₂-to>2P₂O₅

0,2----0,25 mol

n _P=\(\dfrac{6,2}{31}\)=0,2 mol

=>V_O2=0,25.22,4=5,6l

3Fe+2O₂-to>Fe₃O₄

0,03---0,02-----0,01 mol

n _Fe3O4=\(\dfrac{2,32}{232}\)=0,01 mol

=>m _Fe=0,03.56=1,68g

=>m _O2=0,02.32=0,64g

\(4K+O_2\rightarrow2K_2O\\ CH_4+2O_2\rightarrow CO_2+2H_2O\\ 3Fe+2O_2\rightarrow Fe_3O_4\\ 4P+5O_2\rightarrow2P_2O_5\)

a)4 K+O₂-to>2K₂O

b)CH₄+2O₂-to>CO₂+2H₂O

c)3Fe+2O₂-to>Fe₃O₄

d) 4P+5O₂-to>2P₂O₅

Bài 3 : 

PTHH :  \(6Fe+4O_2\left(t^o\right)->2Fe_3O_4\)      (1)

\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{2,32}{56.3+16.4}=0,01\left(mol\right)\)

Từ (1) => \(3n_{Fe_3O_4}=n_{Fe}=0,03\left(mol\right)\)

=> \(m_{Fe}=n.M=1,68\left(g\right)\)

Từ (1) => \(2n_{Fe_3O_4}=n_{O_2}=0,02\left(mol\right)\)

=> \(V_{O_2\left(đktc\right)}=n.22,4=0,448\left(l\right)\)

Bài 4 : 

PTHH : \(4P+5O_2\left(t^o\right)->2P_2O_5\)    (1)

\(n_P=\dfrac{m}{M}=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{32}=0,21\left(mol\right)\)

Có : \(n_P< n_{O_2}\left(0,2< 0,21\right)\)

-> P hết ; O₂ dư

Từ (1) -> \(\dfrac{1}{2}n_P=n_{P_2O_5}=0,1\left(mol\right)\)

=> \(m_{P_2O_5}=n.M=14,2\left(g\right)\)

Bài 3:

\(n_{Fe_3O_4}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

PTHH: 3Fe + 2O₂ ---t^o→ Fe₃O₄

Mol:    0,03     0,02            0,01

\(m_{Fe}=0,03.56=1,68\left(g\right);V_{O_2}=0,02.22,4=0,448\left(l\right)\)

3Fe+2O₂-to>Fe₃O₄

0,03-----0,02-------0,01 mol

n F_e3O4=\(\dfrac{2,32}{232}\)=0,01 mol

=>m _Fe=0,03.56=1,68g

=>V O₂=0,2.22,4=0,448l

bài 2

4P+5O₂-to>2P₂O₅

  0,2---------------0,1

n _P=\(\dfrac{6,2}{31}\)=0,2 mol

n _O2=\(\dfrac{6,72}{22,4}\)=0,3 mol

=>O2 dư , P hết

=>m _P2O5= 0,1.142=14,2g

PTHH : \(4Fe+3O_2\left(t^o\right)-->2Fe_2O_3\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{16.8}{56}=0.3\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22.4}=\dfrac{17.92}{22.4}=0.8\left(mol\right)\)

Có \(n_{Fe}< n_{O_2}\)  (0.3 < 0.8) => O2 dư , Fe hết

\(n_{O_2\left(dư\right)}=n_{O_2\left(PƯ\right)}-n_{Fe}=0.8-0.3=0.5\left(mol\right)\)

=> \(m_{O_2\left(dư\right)}=n_{O_2\left(dư\right)}.M=16\left(g\right)\)

Sản phẩm thu đc lak Fe₂O₃ 

Từ PTHH =>  \(\dfrac{1}{2}n_{Fe}=n_{Fe_2O_3}=0.15\left(mol\right)\)

=> \(m_{Fe_2O_3}=n.M=0,15.\left(56.2+16.3\right)=24\left(g\right)\)

a) \(n_{CH_4}=\dfrac{3,2}{16}=0,2\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

             0,2--->0,4-------->0,2

=> V_O2 = 0,4.22,4 = 8,96(l)

b) m_CO2 = 0,2.44 = 8,8(g)

a, Số mol metan là 

n = m/M = 3,2/16 = 0,2 ( mol ) 

PTHH : CH₄ + 2O₂ -t⁰-> CO₂ + 2H₂O 

1 2 1 2 

0,2 mol -> 0,4 mol 0,2 mol 0,4 mol 

a, Thể tích O₂ ở đktc là : 

V = n . 22,4 = 8,96 (l) 

b, Khối lượng CO₂ là : 

m = n.M = 0,2 . 44 = 8,8 (g) 

Tóm tắt 

mCH4 =3,2(g)

a) Vo=?( lít )

b)mCO2 =?(g)

                       giải 

MCH4=(12+1×4)=16

nCH4=mCH4÷MCH4

          =3,2÷16=0,2( mol )

a)Vo=no×22,4

        =0,2×22,4=4,48( lít )

b) mco2=nco2×Mco2

              =0,2×(12+16×2)

              =0,2×44=8,8(g)