Theo PTHH: \(n_{CaCO_3}=n_{CaO}=n_{CO_2}\)

a) \(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)=n_{CaO}=n_{CO_2}\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,3\cdot56=16,8\left(g\right)\\V_{CO_2}=0,3\cdot22,4=6,72\left(l\right)\end{matrix}\right.\)

b) Tương tự câu a

c) \(n_{CO_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)=n_{CaO}=n_{CaCO_3}\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCO_3}=1,5\cdot100=150\left(g\right)\\m_{CaO}=1,5\cdot56=84\left(g\right)\end{matrix}\right.\)

a. \(CaCO_3\rightarrow CaO+CO_2\)

\(n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)

- Cứ 1 mol CaCO₃ sẽ tạo ra 1 mol CaO và 1 mol CO₂.

=> 0,3 mol CaCO₃ sẽ tạo ra 0,3 mol CaO và 0,3 mol CO₂.

\(m_{CaO}=0,3.56=16,8\left(g\right)\)

\(V_{CO_2}=0,3.22,4=6,72\left(l\right)\)

\(M_{Fe_2O_3}=56.2+16.3=160\left(DvC\right)\)

\(\%Fe=\dfrac{56.2}{160}.100\%=70\%\\ \%O=100\%-70\%=30\%\)

\(M_{SO_2}=32+16.2=64\left(DvC\right)\)

\(\%S=\dfrac{32}{64}.100\%=50\%\\ \%O=100\%-50\%=50\%\)

\(M_{KHCO_3}=39+1+12+16.3=100\left(DvC\right)\)

\(\%K=\dfrac{39}{100}.100\%=39\%\\ \%H=\dfrac{1}{100}.100\%=1\%\\ \%C=\dfrac{12}{100}.100\%=12\%\\ \%O=100\%-39\%-1\%-12\%=48\%\)

Khối lượng của hợp chất Fe₂O₃ là:

M_{\(Fe_2O_3\)}= 56 . 2 + 16 . 3 = 160

%m_Fe = \(\dfrac{112.100\%}{160}=70\%\)

%m_O = 100% - 70% = 30%

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%m_S = \(\dfrac{32.100\%}{64}=50\%\)

%m_O = 100 % - 50% = 50%

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%m_K = \(\dfrac{39.100\%}{100}=39\%\)

%m_H = \(\dfrac{1.100\%}{100}=1\%\)

%m_C = \(\dfrac{12.100\%}{100}=12\%\)

%m_O = 100% - 39% - 1% - 12% = 48%

\(4Fe\left(OH\right)_2+2H_2O+O_2\rightarrow4Fe\left(OH\right)_3\)

\(2Al+3CuO\xrightarrow[]{t^o}Al_2O_3+3Cu\)

\(3Na_2O+2H_3PO_4\rightarrow2Na_3PO_4+3H_2O\)

\(CaO+2HNO_3\rightarrow Ca\left(NO_3\right)_2+H_2O\)

PTHH: \(4A+O_2\xrightarrow[]{t^o}2A_2O\)

Theo PTHH: \(n_A=2n_{A_2O}\) \(\Rightarrow\dfrac{4,6}{M_A}=2\cdot\dfrac{6,2}{2M_A+16}\) \(\Rightarrow M_A=23\)

  Vậy A là Natri

6,2 gam oxit chứ?

1.CuS

2. P₂O₅

3. AlCl₃

4.Zn(NO₃)₂

5.Fe(OH)₃

1. CTHH: Cu_xS_y

\(\dfrac{x}{y}=\dfrac{II}{II}=\dfrac{2}{2}\)

=> CTHH: CuS

2. CTHH: P_xO_y

\(\dfrac{x}{y}=\dfrac{II}{V}=\dfrac{2}{5}\)

=> CTHH: P₂O₅

3. CTHH: Al_xCl_y

\(\dfrac{x}{y}=\dfrac{I}{III}=\dfrac{1}{3}\)

=> CTHH: AlCl₃

4. CTHH: Zn_x(NO₃)_y

\(\dfrac{x}{y}=\dfrac{I}{II}=\dfrac{1}{2}\)

=> CTHH: Zn(NO₃)₂

5. CTHH: Fe_x(OH)_y

\(\dfrac{x}{y}=\dfrac{I}{III}=\dfrac{1}{3}\)

=> CTHH: Fe(OH)₃

\(m_{Al}=0,2.27=5,4\) ( g )

\(m_{Fe}=0,25.56=14\) ( g )

\(m_{Zn}=0,5.65=32,5\) ( g )

=> \(m_{hh_A}=5,4+14+32,5=51,9\) ( g )

a, PTHH: 4P + 5O₂ \(\rightarrow\) 2P₂O₅

b, Theo ĐLBTKL, ta có:

m_P + m_{O\(_2\)} = m_{\(P_2O_5\)}

=> m_{\(O_2\)} = 28,4 - 5,4 = 23 ( g )

\(a\)) \(Al_4C_3+6O_2\overset{t^o}{\rightarrow}3CO_2+2Al_2O_3\)

\(b\)) \(2K_2S+3O_2\overset{t^o}{\rightarrow}2K_2O+2SO_2\)

\(c\)) \(3H_2SO_4+2Al\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

\(d\)) \(2Pb\left(NO_3\right)_2\overset{t^o}{\rightarrow}4NO_2\uparrow+2PbO+4O_2\uparrow\)

ghê thật

\(n_M=\dfrac{11,2}{M}\left(mol\right),n_{HCl}=0,4\left(mol\right)\)

 \(M+2HCl\rightarrow MCl_2+H_2\) (1)

Từ (1)\(\Rightarrow2n_M=n_{HCl}\Leftrightarrow2.\dfrac{11,2}{M}=0,4\)

\(\Leftrightarrow M=56\)

Vậy M là Fe

\(M_{Fe_2O_3}=56.2+16.3=160\left(DvC\right)\)

\(\%Fe=\dfrac{56.2}{160}.100\%=70\%\\ \%O=100\%-70\%=30\%\)