a) Gọi \(\left\{{}\begin{matrix}n_{H_2}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+44b=4,8\\a+b=\dfrac{6,72}{22,4}=0,3\end{matrix}\right.\)

=> a = 0,2; b = 0,1

=> \(\left\{{}\begin{matrix}\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,3}.100\%=66,67\%\\\%V_{CO_2}=100\%-66,67\%=33,33\%\end{matrix}\right.\\\left\{{}\begin{matrix}\%m_{H_2}=\dfrac{0,2.2}{4,8}.100\%=8,33\%\\\%m_{CO_2}=100\%-8,33\%=91,67\%\end{matrix}\right.\end{matrix}\right.\)

b) \(M_{TB}=\dfrac{4,8}{0,3}=16\left(g/mol\right)\)

\(2C_4H_{10}+13O_2->8CO_2+10H_2O\)

\(2C_4H_{10}+13O_2\overset{t^o}{--->}8CO_2+10H_2O\)

\(n_O=\dfrac{12}{16}=0,75\left(mol\right)\\ Số.nguyên.tử.O=0,75.6.10^{23}=4,5.10^{23}\left(nguyên.tử\right)\)

\(n_{O_2}=\dfrac{m}{M}=\dfrac{12}{32}=0,375\left(mol\right)\)

số nguyên tử là

\(0,375\cdot6\cdot10^{23}=2,25\cdot10^{23}\)

\(0,25\cdot6\cdot10^{23}=1,5\cdot10^{23}\\ =>D\)

\(=>D\)

\(M_{hh}=18.2=36\left(g/mol\right);n_{hh}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

=> \(m_{hh}=0,2.36=7,2\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CO}=a\left(mol\right)\\n_{CO_2}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}28a+44b=7,2\\a+b=0,2\end{matrix}\right.\Leftrightarrow a=b=0,1\)

Vậy \(n_{CO}=n_{CO_2}=0,1\left(mol\right)\)

chụp thẳng đc ko chứ mỏi cổ quá

+ CuCl₂ có M = 64 + 35,5.2 = 135 (g/mol)

+ Zn(NO₃)₂ có M = 65 + (14 + 16.3).2 = 189 (g/mol)

Câu 1:

a) `Al_2O_3 + 6HNO_3 -> 2Al(NO_3)_3 + 3H_2O`

b) `3NaOH + Fe(NO_3)_3 -> Fe(OH)_3 + 3NaNO_3`

c) `3CO + Fe_2O_3 -> (t^o) 2Fe + 3CO_2`

d) `MgO + 2HCl -> MgCl_2 + H_2O`

Câu 2:

a) `4P + 5O_2 -> (t^o) 2P_2O_5`

b) `4H_2 + Fe_3O_4 -> (t^o) 3Fe + 4H_2O`

c) `Fe + CuSO_4 -> FeSO_4 + Cu`

d) `Zn + H_2SO_4 -> ZnSO_4 + H_2`

Câu 3:

a) `4Al + 3O_2 -> (t^o) 2Al_2O_3`

b) ĐLBTKL: `m_{Al} + m_{O_2} = m_{Al_2O_3}`

`=> m_{O_2} = 51 - 27 = 24 (g)`

1. a) Al₂O₃ + 6HNO₃ → 2Al(NO₃)₃ + 3H₂O

    b) 3NaOH + Fe(NO₃)₃ → Fe(OH)₃ + 3NaNO₃ 

    c) 3CO + Fe₂O₃ → 2Fe + 3CO₂ 

    d) MgO + 2HCl → MgCl₂ + H₂O 

2. a) 4P + 5O₂ → 2P₂O₅ 

    b) 4H₂ + Fe₂O₄ → 2Fe + 4H₂O

    c) Fe + CuSO₄ → FeSO₄ + Cu

    d) Zn + H₂SO₄ → ZnSO₄ + H₂

3.

a) 

b) ĐLBTKL: 

= 51 − 27 = 24(g)

\(n_{C_{12}H_{22}O_{11}}=\dfrac{32,4}{342}=\dfrac{9}{95}\left(mol\right)\)

⇒ Số phân tử C₁₂H₂₂O₁₁: \(n_{C_{12}H_{22}O_{11}}\cdot N_A=\dfrac{9}{95}\cdot6,022\cdot10^{23}=5,705\cdot10^{23}\) (phân tử)

ai đó giúp mik với 😥