a) nX= 2,24/22,4= 0,1(mol)

=> M(X)= mX/ nX= 2,8/0,1= 28(g/mol)

b) M(B)= mB/nB= 3,6/ 0,15= 24(g/mol)

c) nkhí = 4,48/22,4= 0,2(mol)

=> M(khí)= m(khí)/ n(khí) = 8,8 / 0,2= 44(g/mol)

a) \(n_X=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow M_X=\dfrac{2,8}{0,1}=28\left(g\right)\)

b) \(M_B=\dfrac{3,6}{0,15}=24\left(g\right)\)

c) \(n_{Khí}=\dfrac{4,48}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow M_{Khí}=\dfrac{8,8}{0,2}=44\left(g\right)\)

\(a,M_R=\dfrac{6}{0,15}=40\left(g/mol\right)\\ b,M_A=\dfrac{m_A}{n_A}=\dfrac{7}{\dfrac{5,6}{22,4}}=\dfrac{7}{0,25}=28\left(g/mol\right)\\ c,\overline{M_{hh}}=\dfrac{4\cdot28+1\cdot32}{4+1}=\dfrac{144}{5}=28,8\left(g/mol\right)\)

Ta có : \(n_{O2}=\dfrac{V}{22,4}=0,05\left(mol\right)\)

\(\Rightarrow O2=n.A=3.10^{22}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{V}{22,4}=0,1\left(mol\right)\)

\(\Rightarrow SO3=n.A=6.10^{22}\) ( phân tử )

Ta có : \(n_{NaOH}=\dfrac{m}{M}=0,4\left(mol\right)\)

\(\Rightarrow NaOH=n.A=2,4.10^{23}\) ( phân tử )

Ta có : \(n_{SO3}=\dfrac{m}{M}=0,405\left(mol\right)\)

\(\Rightarrow SO3=n.A=2,4381.10^{23}\) ( phân tử )

\(V_{CO_2}=0,5.22,4=11,2\left(l\right)\)

\(A_{CO_2}=0,5.6.10^{23}=3.10^{23}\) (phân tử \(CO_2\) )

2.

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

=> \(n_C=n_{CO_2}=0,1\left(mol\right)\) (1)

=> \(n_O=2nCO_2=0,1.2=0,2\left(mol\right)\) (*)

\(n_{H_2O}=\dfrac{3,6}{18}=0,2\left(mol\right)\)

=> \(n_H=2n_{H_2O}=0,2.2=0,4\left(mol\right)\) (2)

=> \(n_O=n_{H_2O}=0,2\left(mol\right)\) (**)

\(n_{O_2}=\dfrac{4,8}{22,4}=0,2\left(mol\right)\)

=> \(n_O=2n_{O_2}=2.0,2=0,4\left(mol\right)\) (3)

\(X+O_2\underrightarrow{t^o}CO_2+H_2O\)

Từ (1),(2),(3), (*), (**)  suy ra: \(n_C:n_H:n_O=0,1:0,4:0\)

=> Công thức tổng quát của X là \(C_xH_y\)

có: \(x:y=n_C:n_H=0,1:0,4=1:4\)

=> X là: \(CH_4\)

Sơ đồ pứ: \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

\(m_{CH_4}=3,6+0,2.44-0,2.32=6\left(g\right)\)

\(d_{\dfrac{A}{B}}=0,5\Rightarrow\dfrac{A}{B}=\dfrac{1}{2}\Leftrightarrow2A=B\)

\(n_B=\dfrac{1}{22,4}=\dfrac{2,857}{B}\Rightarrow B=2,857.22,4\approx64\left(g\right)\)

=> A= 64/2 = 32(g)

vậy M_A= 32(g)

       M_B= 64(g)

1)

a) \(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)

b) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

c) \(n_{H_2O}=\dfrac{15.10^{23}}{6.10^{23}}=2,5\left(mol\right)\)

2)

a) \(n_A=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) => M_A = \(\dfrac{3}{0,1}=30\left(g/mol\right)\)

b) \(d_{A/O_2}=\dfrac{30}{32}=0,9375\)

Bài 1 : 

Số mol , khối lượng , số phân tử của các chất lần lượt là : 

\(a.\)\(\)

\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(m_{O_2}=0.05\cdot32=1.6\left(g\right)\)

\(0.05\cdot6\cdot10^{23}=0.3\cdot10^{23}\left(pt\right)\)

\(b.\)

\(n_{SO_3}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(m_{SO_3}=0.1\cdot80=8\left(g\right)\)

\(0.1\cdot6\cdot10^{23}=0.6\cdot10^{23}\left(pt\right)\)

\(c.\)

\(n_{H_2S}=\dfrac{36}{22.4}=\dfrac{45}{28}\left(mol\right)\)

\(m_{H_2S}=\dfrac{45}{28}\cdot34=\dfrac{765}{14}\left(g\right)\)

\(\dfrac{45}{28}\cdot6\cdot10^{23}=\dfrac{135}{14}\cdot10^{23}\left(pt\right)\)

\(d.\)

\(n_{C_4H_{10}}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(m_{C_4H_{10}}=0.2\cdot58=11.6\left(g\right)\)

\(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

Bài 2 : 

\(a.\)

\(n_{SO_3}=\dfrac{16}{80}=0.2\left(mol\right)\)

Số phân tử SO₃ : \(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

\(b.\)

\(n_{NaOH}=\dfrac{8}{40}=0.2\left(mol\right)\)

Số phân tử NaOH : \(0.2\cdot6\cdot10^{23}=1.2\cdot10^{23}\left(pt\right)\)

\(c.\)

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{16}{400}=0.04\left(mol\right)\)

Số phân tử Fe₂(SO₄)₃ : \(0.04\cdot6\cdot10^{23}=0.24\cdot10^{23}\left(pt\right)\)

\(d.\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{34.2}{342}=0.1\left(mol\right)\)

Số phân tử Al₂(SO₄)₃ : \(0.1\cdot6\cdot10^{23}=0.6\cdot10^{23}\left(pt\right)\)

Bài 2 Tính số mol , số phân tử các chất sau:

a, 16 gam SO3

\(n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\)

b, 8 gam NaOH

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

c, 16 gam Fe2(SO4)3

\(n_{Fe_2\left(SO_4\right)_3}=\dfrac{16}{400}=0,04\left(mol\right)\)

d, 34,2 gam Al2(SO4)3

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{342}=0,1\left(mol\right)\)

a . mA= 0,5x32=16 g

b . n=36x10²³/6x10²³=6(mol)

V=6x22,4=134,4 g