Số \(mol\) của \(36g\) \(CH_4:\)

\(n_{CH_4}=\dfrac{m}{M}=\dfrac{36}{16}=2,25\left(mol\right)\)

Trong \(CH_4\) có \(:\)

\(-4mol\) \(H\)

\(\Rightarrow n_H=4\times n_{CH_4}=4\times2,25=9\left(mol\right)\)

TK:

https://lazi.vn/edu/exercise/452918/dot-chay-16g-chat-a-can-4-48-lit-khi-oxi-o-dktc-thu-duoc-khi-co2-va-hoi-nuoc-theo-ti-le-so-mol-la-1-2-tinh-khoi-luong

Gọi CTPT A là C_xH_yO_z

PTHH: 2C_xH_yO_z + \(\dfrac{4x+y-2z}{2}\)O₂ --t^o--> 2xCO₂ + yH₂O 

______1------->\(\dfrac{4x+y-2z}{4}\)------------->x-------->0,5y

=> \(\left\{{}\begin{matrix}\dfrac{4x+y-2z}{4}=6,5\\x=4\\0,5y=5\end{matrix}\right.=>\left\{{}\begin{matrix}x=4\\y=10\\z=0\end{matrix}\right.\)

=> CTPT: C₄H₁₀

TK: 

https://hoidap247.com/cau-hoi/113717

\(n_{H_2SO_4}=\dfrac{7,35}{98}=0,075\left(mol\right)\)

PTHH: A + H₂SO₄ --> ASO₄ + H₂

____0,06<-0,06

2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

_0,03-->0,015

=> \(M_A=\dfrac{1,44}{0,06}=24\left(Mg\right)\)

63 : 9 = 7

HT

63 : 9 = 7

a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

____\(\dfrac{4}{15}\)<----0,4<--------------------0,4

=> \(m_{Al}=\dfrac{4}{15}.27=7,2\left(g\right)\)

c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,4}{0,15}=2,667M\)

Có n_N : n_O = 1:2

=> CTHH: (NO₂)_n

Mà PTK_A = 92(đvC)

=> n = 2

=> CTHH: N₂O₄

Do 1 lít khí B nặng bằng 1 lít khí CO₂

=> PTK_B = PTK_CO2 = 44

CTHH: N_xO_y

- Nếu x = 1 => L

- Nếu x = 2 => y = 1 

=> CTHH: N₂O

Vậy A là N₂O₄, B là N₂O