B1 sửa 4,69 gam -> 4,6 gam

\(B1\\ n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\\ 2R+2H_2O\rightarrow2ROH+H_2\\ n_R=2.n_{H_2}=2.0,1=0,2\left(mol\right)\\ M_R=\dfrac{4,6}{0,2}=23\left(\dfrac{g}{mol}\right)\)

=> R(I) là Natri (Na=23)

R + H2O -> ROH + 1/2 H2

nH2= 0,15(mol)

=> nROH=0,3(mol)

mROH= 6%.200=12(g)

=> M(ROH)= 12/0,3=40(g/mol)

Mà: M(ROH)=M(R)+17

=>M(R)+17=40

=>M(R)=23(g/mol) => R là Natri (Na=23)

Đáp án B

Đáp án B

Đáp án B

=> R = 23n

\(R+2HCl\rightarrow RCl_2+H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow n_R=n_{H_2}=0,1\left(mol\right)\\ M_R=\dfrac{5,6}{0,1}=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(II\right):Sắt\left(Fe=56\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\)

Bài 1:

Gọi KL cần tìm là A.
PT: \(A+2HCl\rightarrow ACl_2+H_2\)

Ta có: \(n_{HCl}=0,1.6=0,6\left(mol\right)\)

Theo PT: \(n_A=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

\(\Rightarrow M_A=\dfrac{7,2}{0,3}=24\left(g/mol\right)\)

Vậy: KL cần tìm là Mg.

Bài 2:

PT: \(2R+6HCl\rightarrow2RCl_3+3H_2\)

Ta có: \(n_{H_2}=\dfrac{9,408}{22,4}=0,42\left(mol\right)\)

Theo PT: \(n_R=\dfrac{2}{3}n_{H_2}=0,28\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{7,56}{0,28}=27\left(g/mol\right)\)

Vậy: R là Al.

\(R+H_2SO_4\rightarrow RSO_4+H_2\\ n_{H_2}=\dfrac{7,168}{22,4}=0,32\left(mol\right)\\ n_R=n_{H_2}=0,32\left(mol\right)\\ M_R=\dfrac{7,68}{0,32}=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(II\right):Magie\left(Mg=24\right)\)