Câu 2:

PTHH: \(2KMnO_4\xrightarrow[]{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)

Ta có: \(n_{KMnO_4}=\dfrac{30,8}{158}=\dfrac{77}{395}\left(mol\right)\)

\(\Rightarrow n_{O_2\left(thực\right)}=\dfrac{\dfrac{77}{395}}{2}\cdot90\%=\dfrac{693}{7900}\left(mol\right)\) \(\Rightarrow V_{O_2}=\dfrac{693}{7900}\cdot22,4\approx1,96\left(l\right)\) 

Câu 1:

PTHH: \(M_2O_3+6HNO_3\rightarrow2M\left(NO_3\right)_3+3H_2O\)

Theo PTHH: \(n_{M_2O_3}=\dfrac{1}{3}n_{HNO_3}=1\left(mol\right)\)

\(\Rightarrow\dfrac{5,1}{2M+16\cdot3}=1\) \(\Rightarrow M< 0\)

  Vậy đề bài sai :) 

\(n_{A_2O}=\dfrac{9,4}{2M_A+16}\left(mol\right)\)

PTHH: A₂O + 2HCl --> 2ACl + H₂O

       \(\dfrac{9,4}{2M_A+16}\)-->\(\dfrac{9,4}{M_A+8}\)

=> \(\dfrac{9,4}{M_A+8}\left(M_A+35,5\right)=14,9\Rightarrow M_A=39\left(g/mol\right)\)

=> A là K

CTHH: K₂O

a> RO + 2HCl ---> RCl2 + H2O

b) Số mol RO = số mol H2O = 0,05 mol.

Do đó: R + 16 = 2,8/0,05 = 56 nên R = 40 (Ca).

Pthh RO + 2HCl --> RCl₂ + H₂O

theo đề 0,05 0,1 0,05 0,05 (mol)

n_H2O=0,9/18=0,05(mol)

m_oxit=2,8/0,05=56 (g)

ta có RO=56(g)

<-> R + 16 = 56

<-> R= 56 - 16 = 40

Vậy R là Canxi

Bài 1:

\(n_M=\dfrac{16}{M_M}\left(mol\right)\)

PTHH: 2M + O₂ --t^o--> 2MO

         \(\dfrac{16}{M_M}\)---------->\(\dfrac{16}{M_M}\)

=> \(\dfrac{16}{M_M}\left(M_M+16\right)=20\)

=> M_M = 64 (g/mol)

=> M là Cu

Bài 2:

\(n_R=\dfrac{16,2}{M_R}\left(mol\right)\)

PTHH: 2R + 3Cl₂ --t^o--> 2RCl₃

          \(\dfrac{16,2}{M_R}\)------------>\(\dfrac{16,2}{M_R}\)

=> \(\dfrac{16,2}{M_R}\left(M_R+106,5\right)=80,1\)

=> M_R = 27 (g/mol)

=> R là Al

 1 
ADDDLBTKL ta có
\(m_{O_2}=m_{MO}-m_M\\ m_{O_2}=20-16=4g\\ n_{O_2}=\dfrac{4}{32}=0,125\left(mol\right)\\ pthh:2M+O_2\underrightarrow{t^o}2MO\) 
            0,25   0,125 
\(M_M=\dfrac{16}{0,25}=64\left(\dfrac{g}{mol}\right)\) 
=> M là Cu 
2 
ADĐLBTKL ta có 
\(m_{Cl_2}=m_{RCl_3}-m_R\\ m_{Cl_2}=80,1-16,2=63,9g\\ n_{Cl_2}=\dfrac{63,9}{71}=0,9\left(mol\right)\\ pthh:2R+3Cl_2\underrightarrow{t^o}2RCl_3\) 
            0,6   0,9 
\(M_R=\dfrac{16,2}{0,6}=27\left(\dfrac{g}{mol}\right)\) 
=> R là Al

\(1 ) 2M+O_2\rightarrow 2MO n_M=n_{MO}\Leftrightarrow \dfrac{16}{M_M}=\dfrac{20}{m_M+16} \Rightarrow m_m = 64(g/mol) \rightarrow M : Cu \)

\(2) 2R+3Cl_2\rightarrow 2RCl_3 n_R=nn_{RCl_3}\Leftrightarrow \dfrac{16,2}{M_R}=\dfrac{80,1}{M_R+35,5.3}\Rightarrow M_R = 27(g/mol)\rightarrow R:Al \)

a/Ta có PTHH:

RO + 2HCl \(\rightarrow\) RCl2 + H2O

b) Mol RO = mol H2O = 0,05 mol.

\(\Rightarrow R+16=\dfrac{2,8}{0,05}=56\)

\(\Rightarrow R=56-16=40\)

Vậy nguyên tử R là Canxi (Ca)

https://hoc24.vn/hoi-dap/question/24299.html

a) PTHH : \(RO+2HCl\rightarrow RCl_2+H_2O\)

b) Cách 1: \(n_{H_2O}=\dfrac{0,9}{18}=0,05\left(mol\right)\)

Mà : \(n_{RO}=n_{H_2O}=0,05mol\)

\(\Rightarrow R+16=\dfrac{2,8}{0,05}\)

\(\Rightarrow R=\dfrac{2,8}{0,05}-16=40\)

Vậy kim loại R là Ca.

Cách 2 : Theo PTHH ta có :

\(n_{RO}=n_{H_2O}\)

\(\Rightarrow\dfrac{m_{RO}}{M_{RO}}=\dfrac{m_{H_2O}}{M_{H_2O}}\)

\(\Rightarrow\dfrac{2,8}{R+16}=\dfrac{0,9}{18}\)

\(\Rightarrow2,8.18=0,9R+14,4\)

\(\Rightarrow R=\dfrac{50,4-14,4}{0,9}=40\)

Vậy kim loại R là Ca.

\(A_2O+2HNO_3\rightarrow2ANO_3+H_2O\)

\(n_{A_2O}=\dfrac{1}{2}n_{HNO_3}=0,2\left(mol\right)\)

=> \(M_{A_2O}=\dfrac{12,4}{0,2}=62\)

Ta có : 2A + 16 =62 

=> A=23 (Na) 

Vậy oxit cần tìm là Na2O

$n_{HNO_3} = 0,2.2 = 0,4(mol)$

Gọi oxit cần tìm là $A_2O$
$A_2O + 2HNO_3 \to 2ANO_3 + H_2O$
$n_{oxit} = \dfrac{1}{2}n_{HNO_3} = 0,2(mol)$
$\Rightarrow M_{oxit} = 2A + 16 = \dfrac{12,4}{0,2} = 62$

$\Rightarrow A = 23(Natri)$