Bài 24: Tính chất của oxi

a) S + O₂ --t^o--> SO₂

b) \(n_S=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

PTHH: S + O₂ --t^o--> SO₂

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,15}{1}\) => S dư, O₂ hết

c) 

PTHH: S + O₂ --t^o--> SO₂

________0,15----->0,15

=> m_SO2 = 0,15.64 = 9,6(g)

\(a,PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ b,n_{Fe}=\dfrac{5,6}{56}=0,1(mol);n_{O_2}=\dfrac{3,2}{32}=0,1(mol)\)

Vì \(\dfrac{n_{Fe}}{3}<\dfrac{n_{O_2}}{2}\) nên \(O_2\) dư

\(n_{O_2(dư)}=0,1-0,1.\dfrac{2}{3}=0,033(mol)\\ \Rightarrow m_{O_2(dư)}=0,033.32=1,056(mol)\\ c,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,033(mol)\\ \Rightarrow m_{Fe_3O_4}=0,033.232=7,656(g)\)

sửa lại hộ mình chỗ \(m_{O_2(dư)}=1,056(g)\)

\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ \left(mol\right).....0,1..........0,1\leftarrow0,15\\ m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

Ta có: \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: \(2KClO_3\xrightarrow[t^o]{V_2O_5}2KCl+3O_2\)

Theo PT: \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)

=> \(m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)

\(CH_4+ZO_2\rightarrow CO_2+2H_2O\)

0,2         0,4        0,2

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

\(\Rightarrow V_{O_2}=8,96l\\ m_{CO_2}=n.M=0,2.44=8,8\left(g\right)\)

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

PTHH: CH₄ + 2O₂ ---t^o→ CO₂ + 2H₂O

Mol:      0,2      0,4

\(V_{O_2}=0,4.22,4=8,96\left(l\right)\)

\(n_{O_2\left(đktc\right)}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ 0,12........0,15.........0,06\left(mol\right)\\ m_P=0,12.31=3,72\left(g\right)\)

Ta có: \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: 4P + 5O₂ ---t^o---> 2P₂O₅.

Theo PT: n_P = \(\dfrac{4}{5}.n_{O_2}=\dfrac{4}{5}.0,15=0,12\left(mol\right)\)

=> m_P = 31 . 0,12 = 3,72(g)

PTHH: \(S+O_2\xrightarrow[]{t^o}SO_2\)

a) Ta có: \(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{S\left(p/ứ\right)}\) 

\(\Rightarrow m_{S\left(p/ứ\right)}=0,1\cdot32=3,2\left(g\right)\) \(\Rightarrow m_{tạp.chất}=3,25-3,2=0,05\left(g\right)\)

b) Ta có: \(\%m_S=\dfrac{3,2}{3,25}\cdot100\%\approx98,46\%\)

Trong hóa học D là là kí hiệu của đơteri, một đồng vị của Hidro (hay ký hiệu là H)

Khối lượng riêng em nha !

a)

$CaO + H_2O \to Ca(OH)_2$

$CO_2 + H_2O \rightleftharpoons H_2CO_3$

$N_2O_5 + H_2O \to 2HNO_3$

b)

$CaO + 2HCl \to CaCl_2 + H_2O$
$Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O$
$MgO + 2HCl \to MgCl_2 + H_2O$

c)

$CO_2 + KOH \to KHCO_3$
$N_2O_5 + 2KOH \to 2KNO_3 + H_2O$