\(^nH_2=\dfrac{5,04}{22,4}=0,225\left(mol\right)\)

Gọi hóa trị của kim loại Z là x

              2Z     +     2xHCl     --->       2\(ZCl_x\)       +         xH\(_2\)

Mol       \(\dfrac{0,45}{x}\)                                                                0,225

--> Z = \(\dfrac{4,05.x}{0,45}\) = 9x

--> x = 3 thì Z = 27 (Al)

Có \(^nAl=\dfrac{5,4}{27}=0,2\left(mol\right)\)

         \(4Al+3O_2\rightarrow2Al_2O_3\)

Mol    0,2     0,15

Có \(\%VO_2\left(kk\right)=20\%\)

--> V không khí cần dùng = \(\dfrac{0,15}{20\%}\) = 0,75 (mol)

Chúc bạn học tốt!!!

\(n_A=\dfrac{5,4}{M_A}\left(mol\right)\)

PTHH: 2xA + yO₂ --t^o--> 2A_xO_y

_____\(\dfrac{5,4}{M_A}\) ------------->\(\dfrac{5,4}{x.M_A}\)

=> \(\dfrac{5,4}{x.M_A}\left(x.M_A+16y\right)=10,2\)

=> \(M_A=9.\dfrac{2y}{x}\)

Xét \(\dfrac{2y}{x}=3=>M_A=27\left(Al\right)=>\dfrac{x}{y}=\dfrac{2}{3}=>Al_2O_3\)

\(2xA+yO_2\overset{t^o}{--->}2A_xO_y\)

Áp dụng ĐLBTKL, ta có:

\(m_A+m_{O_2}=m_{A_xO_y}\)

\(\Leftrightarrow5,4+m_{O_2}=10,2\)

\(\Leftrightarrow m_{O_2}=10,2-5,4=4,8\left(g\right)\)

\(\Rightarrow n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

Theo PT: \(n_{A_xO_y}=\dfrac{2}{y}.n_{O_2}=\dfrac{2}{y}.0,15=\dfrac{0,3}{y}\left(mol\right)\)

\(\Rightarrow n_{A_{\left(A_xO_y\right)}}=\dfrac{0,3}{y}.x=\dfrac{0,3x}{y}\left(mol\right)\)

Theo PT: \(n_A=\dfrac{2x}{y}.n_{O_2}=\dfrac{0,3x}{y}\left(mol\right)\)

\(\Rightarrow\dfrac{0,3x}{y}.A=5,4\)

\(\Leftrightarrow\dfrac{A}{9}=\dfrac{2y}{x}\)

Biện luận:

Vậy A là nhôm (Al)

nH2 = 6,72/22,4 = 0,3 (mol)

PTHH: 2A + 6HCl -> 2ACl3 + 3H2

nA = 0,3 : 3 . 2 = 0,2 (mol)

M(A) = 5,4/0,2 = 27 (g/mol)

A là nhôm Al

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(PTHH:2A+6HCl\text{⟶2ACl3}+3H_2\)

\(n_A=0,3:3.2=0,2\left(mol\right)\)

\(M_A=\dfrac{5,4}{0,2}=27\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow A=Al\)

Ta có: \(n_{H_2}=\dfrac{0,224}{22,4}=0,01\left(mol\right)\)

BT e, có: 2n_R = 2n_H2 ⇒ n_R = 0,01 (mol)

\(\Rightarrow M_R=\dfrac{1,37}{0,01}=137\left(g/mol\right)\)

→ R là Ba.

Đáp án B

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ KL:M\\ M+2HCl\rightarrow MCl_2+H_2\\ n_{MCl_2}=n_M=n_{H_2}=0,05\left(mol\right)\\ M_{MCl_2}=\dfrac{4,75}{0,05}=95\left(\dfrac{g}{mol}\right)\\ M\text{à}:M_{MCl_2}=M_M+71\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_M+71=95\\ \Leftrightarrow M_M=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow M\left(II\right):Magie\left(Mg=24\right)\\ a=24.0,05=1,2\left(g\right)\)

Bạn xem lại đề gõ thật chuẩn lại nhé!

Bài 2:  \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PTHH: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\)

Theo PTHH: \(n_{FeCl_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=127\cdot0,1=12,7\left(g\right)\)