\(PTHH:C+O_2->CO_2\)

ap dung DLBTKL ta co

\(m_C+m_{O_2}=m_{CO_2}\\ m_{O_2}=m_{CO_2}-m_C\\ m_{O_2}=88-24\\ m_{O_2}=64\left(g\right)\)

\(PTHH:S+O_2->SO_2\)

ap dung DLBTKL ta co

\(m_S+m_{O_2}=m_{SO_2}\)

\(=>m_{O_2}=m_{SO_2}-m_S\\ =>m_{O_2}=96-48\\ =>m_{O_2}=48\left(g\right)\)

thank

\(4Al+3O_2->2Al_2O_3\)

1. \(ZnCl\left(I\right)_2=x . 1=I . 2=>x=\dfrac{I . 2}{1}=II\) 
=> Zn hóa trị II
2. \(AlCl\left(I\right)_3=x . 1=I . 3=>x=\dfrac{I . 3}{1}=III\) 
=> Al hóa trị III
3. \(CuCl\left(I\right)_2=x . 1=I . 2=>x=\dfrac{I . 2}{1}=II\) 
=> Cu hóa trị II
4. \(Fe_2O\left(II\right)_3=x . 2=II . 3=>x=\dfrac{II . 3}{2}=III\) 
=> Fe hóa trị III

1: Hóa trị II

2: Hóa trị III

3: Hóa trị II

4; Hóa trị III

a) \(n_{H_2O}=\dfrac{1,8}{18}=0,1\left(mol\right)\)

b) \(m_{NaCl}=0,2.58,5=11,7\left(g\right)\)

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

d) \(n_{CO_2}=\dfrac{0,6.10^{23}}{6.10^{23}}=0,1\left(mol\right)\Rightarrow m_{CO_2}=0,1.44=4,4\left(g\right)\)

Ta có: \(\%m_O=\dfrac{16y}{14x+16y}.100\%=69,6\%\)

=> \(\dfrac{x}{y}=\dfrac{1}{2}\)

=> N_xO_y có là N_xO_2x hay (NO₂)_x

Giả sử có 1 mol khí

=> \(\left\{{}\begin{matrix}n_{NO}=0,45\left(mol\right)\\n_{NO_2}=0,15\left(mol\right)\\n_{\left(NO_2\right)_x}=0,4\left(mol\right)\end{matrix}\right.\)

=> \(\%m_{NO}=\dfrac{0,45.30}{0,45.30+0,15.46+0,4.46x}.100\%=23,6\%\)

=> \(x=2\)

=> N_xO_y là N₂O₄

a) Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)

=> 27a + 65b = 15,16 (1)
PTHH:

4Al + 3O₂ --t^o--> 2Al₂O₃ 

a---------------->0,5a

2Zn + O₂ --t^o--> 2ZnO

b------------------>b

=> 102.0,5a + 81b = 51a + 81b = 20,28 (2)

Từ (1), (2) => a = 0,08; b = 0,2

=> \(\left\{{}\begin{matrix}m_{Al_2O_3}=102.0,5.0,08=4,08\left(g\right)\\m_{ZnO}=0,2.81=16,2\left(g\right)\end{matrix}\right.\)

b) ĐLBTKL: \(m_{O_2}=20,28-15,16=5,12\left(g\right)\)

=> \(n_{O_2}=\dfrac{5,12}{32}=0,16\left(mol\right)\)

=> \(V_{kk}=\dfrac{0,16.22,4}{20\%}=17,92\left(l\right)\)

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