a) M_X = 2.23 = 46(g/mol)

b) \(m_N=\dfrac{46.30,43}{100}=14\left(g\right)=>n_N=\dfrac{14}{14}=1\left(mol\right)\)

\(m_O=46-14=32\left(g\right)=>n_O=\dfrac{32}{16}=2\left(mol\right)\)

=> CTHH: NO₂

\(a.M_X=23M_{H_2}=46\left(g/mol\right)\\ b.ĐặtCT:N_xO_y\\ Tacó:\%N=\dfrac{14x}{64}=30,43\%\\ \Rightarrow x=1\\ Tacó:14.1+y.16=46\\ \Rightarrow y=2\\ VậyCT:NO_2\)

\(a,M_X=23.M_{H_2}=23.2=46(g/mol)\\ b,\text{Đặt }CTHH_X:N_xO_y\\ \Rightarrow x:y=\dfrac{30,43}{14}:\dfrac{100-30,43}{16}\approx \dfrac{1}{2}\\ \Rightarrow x=1;y=2\\ \Rightarrow CTHH:N_2O\)

1)

$M_X = 1,375.32  = 44(g/mol)$
$M_X = 0,0625.32 = 2(g/mol)$

2)

$M_X = 2,207.29 = 64(g/mol)$

$M_X = 1,172.29 = 34(g/mol)$

3)

$M_X = 17.2 = 34(g/mol)$

Vậy khí X là $H_2S$

4)

a) $M_X = 0,552.29 = 16$
Gọi CTHH của X là $C_xH_y$

Ta có :  $\dfrac{12x}{75} = \dfrac{y}{25} = \dfrac{16}{100}$

Suy ra:  x = 1 ; y = 4

Vậy X là $CH_4$

$CH_4 + 2O_2 \xrightarrow{t^o} CO_ 2+ 2H_2O$
$V_{O_2} = 2V_{CH_4} = 11,2.2 = 22,4(lít)$

\(a,M_A=1,8125.32=58(g/mol)\)

Trong 1 mol A: \(\begin{cases} n_C=\dfrac{58.82,75\%}{12}=4(mol)\\ n_H=\dfrac{58-4.12}{1}=10(mol) \end{cases}\)

\(\Rightarrow CTHH_A:C_4H_{10}\\ b,n_C=\dfrac{2,88}{12}=0,24(mol)\\ \Rightarrow n_A=\dfrac{1}{4}n_C=0,06(mol)\\ \Rightarrow m_A=0,06.58=3,48(g)\\ c,n_A=\dfrac{11,6}{58}=0,2(mol)\\ \Rightarrow \begin{cases} n_C=0,2.4=0,8(mol)\\ n_H=0,2.10=2(mol) \end{cases}\Rightarrow \begin{cases} m_C=0,8.12=9,6(g)\\ m_H=2.1=2(g) \end{cases}\)

a,Ta có:M_X = 23.2 = 46 (g/mol)

b,Gọi CTHH của X là N_yO_z

=> m_N = 46.30,43%=14 (g)=>y=14:14=1

=> m_O = 46-14 = 32 (g) => z=32:16=2

Vậy CTHH: NO₂

Gọi CTTQ là : XO₃ 

\(a,\rightarrow M_A=80\)

\(\rightarrow M_X=80-\left(16.3\right)=32\left(\dfrac{g}{mol}\right)\)

Vậy X là S ( lưu huỳnh )

\(\%m_{S\left(SO_3\right)}=\dfrac{32}{80}.100\%=40\%\)

\(d_{\dfrac{A}{H_2}}=\dfrac{M_A}{M_{H_2}}=40\Rightarrow M_A=40.M_{H_2}=40.2=80\) (g/mol)

\(m_O=80.\dfrac{60}{100}=48\left(g\right)\)

\(m_S=80.\dfrac{40}{100}=32\left(g\right)\)

\(n_O=\dfrac{m}{M}=\dfrac{48}{16}=3\left(mol\right)\)

\(n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\)

\(\Rightarrow\)CTHH của khí A là SO₃

TK