Bài 24: Tính chất của oxi

Đáp án A

đáp án B:88,9%

%Oxi có trong nước là bao nhiêu ? A. 11.1%. B. 88.9%. C. 90%. D. 10%

a. 4P + 5O2 -> (nhiệt độ) 2P2O5.

b. nP2O5 = 7,1/142 = 0,05 mol

=> nO2 pư = 0,125 mol

=> VO2 thực tế = 0,125 x 1,15 x 22,4 = 3,22 L

1. \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(2Mg+O_2\xrightarrow[]{t^\circ}2MgO\)

 0,2 → 0,1

\(\Rightarrow V_{O_2}=0,1\cdot22,4=2,24\left(l\right)\)

2. \(n_{P_2O_5}=\dfrac{28,2}{142}=\dfrac{141}{710}\left(mol\right)\)

\(4P+5O_2\xrightarrow[]{t^\circ}2P_2O_5\)

\(\dfrac{141}{355}\)       ←       \(\dfrac{141}{710}\)

\(\Rightarrow m_P=\dfrac{141}{355}\cdot31\approx12,312\left(g\right)\)

1, \(2Mg+O_2--->2MgO\) 

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

Theo pt \(n_{O_2}=\dfrac{n_{Mg}}{2}=\dfrac{0,2}{2}=0,1\left(mol\right)\)

\(V_{O_2}=n\times22,4=0,1\times22,4=2,24\left(lít\right)\)

2, \(4P+5O_2-->2P_2O_5\)

\(n_{P_2O_5}=\dfrac{28,2}{2\times31+5\times16}\approx0,2\left(mol\right)\)

Theo phương trình :

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

\(m_P=0,1\times31=3,1\left(g\right)\)

Bài 1:

PT: \(4Na+O_2\underrightarrow{t^o}2Na_2O\)

\(Ca+\dfrac{1}{2}O_2\underrightarrow{t^o}CaO\)

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

\(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)

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

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

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

\(C_2H_6+\dfrac{7}{2}O_2\underrightarrow{t^o}2CO_2+3H_2O\)

\(C_6H_6+\dfrac{15}{2}O_2\underrightarrow{t^o}6CO_2+3H_2O\)

Bài 2:

Ta có: \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

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

Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\) \(\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

Bài 3:

Ta có: \(n_{C_2H_2}=\dfrac{2,6}{26}=0,1\left(mol\right)\)

PT: \(2C_2H_2+5O_2\underrightarrow{t^o}4CO_2+2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{5}{2}n_{C_2H_2}=0,25\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,25.24,79=6,1975\left(l\right)\)

a)

$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$

b) Theo PTHH : $V_{O_2} = 2V_{CH_4} = 11,2(lít)$

$n_{CH_4} = \dfrac{5,6}{22,4} = 0,25(mol)$

Theo PTHH : $n_{H_2O} = 2n_{CH_4} = 0,5(mol)$
$m_{H_2O} = 0,5.18 = 9(gam)$

\(a,CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(1:2:1:2\left(mol\right)\)

\(0,25:0,5:0,25:0,5\left(mol\right)\)

\(n_{CH_4}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,5.22,4=11,2\left(l\right)\)

\(m_{H_2O}=n.M=0,5.18=9\left(g\right)\)

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

Ta có: \(n_{O_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

a, Theo PT: \(n_C=n_{O_2}=0,025\left(mol\right)\)

\(\Rightarrow m_C=0,025.12=0,3\left(g\right)\)

b, Theo PT: \(n_{CO_2}=n_{O_2}=0,025\left(mol\right)\)

\(\Rightarrow V_{CO_2}=0,025.22,4=0,56\left(l\right)\)

$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$P_2O_5 + 3H_2O \to 2H_3PO_4$
$4Na + O_2 \xrightarrow{t^o} 2Na_2O$
$Na_2O + H_2O \to 2NaOH$
$2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
$2H_2O \xrightarrow{đp} 2H_2 + O_2$

$S + O_2 \xrightarrow{t^o} SO_2$
$2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$SO_3 + H_2O \to H_2SO_4$

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

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

             0,1-------------->0,05

=> m_P2O5 = 0,05.142 = 7,1 (g)

=> A

nP = 12,4 :31 = 0,4 (mol) 
nO2 = 12,8 : 32= 0,4 (mol) 
pthh: 4P + 5O2 -t-> 2P2O5 
LTL : \(\dfrac{0,4}{4}\) < \(\dfrac{0,4}{5}\)
=> P dư 
theo pthh , nP(pư)= \(\dfrac{4}{5}\)nO2 = 0,32 (mol) 
=> nP(d) = nO2(bđ) - nO2(pư ) = 0,4 - 0,32 = 0,08 (mol) 
=> mP(d) = 0,08 . 31 = 2,48 (G)

bn tham khảo