`a)`

`2Mg(NO_3)_2` $\xrightarrow{t^o}$ `2MgO+4NO_2+O_2`

`b)`

`n_{Mg(NO_3)_2}={33,5}/{148}≈0,226(mol)`

Theo pt:

`n_{MgO}=n_{Mg(NO_3)_2}=0,226(mol)`

`n_{O_2}=1/2n_{MgO}=0,113(mol)`

`m_{MgO}=0,226.40=9,04(g)`

`V_{O_2}=0,113.24,79=2,80127(L)`

`c)`

`n_{NO_2}+n_{O_2}={61,975}/{24,79}=2,5(mol)`

Theo pt:

`n_{NO_2}=4n_{O_2}`

`->n_{NO_2}=2(mol);n_{O_2}=0,5(mol)`

Theo pt:

`n_{Mg(NO_3)_2\  t t }=2n_{O_2\ t t}=1(mol)`

`->m_{Mg(NO_3)_2\ l t}={1.148}/{70\%}≈211,43(g)`

a) 2Mg(NO3)2 -> 2MgO + 4NO2 + O2

b) n Mg(NO3)2 = 33,5/148= 67/296

     2Mg(NO3)2 -> 2MgO + 4NO2 + O2

      67/296 ->       67/296                 67/592

     => m MgO = 67/296 . 40 = 9,054 (g)

           V O2 = 67/592 . 24,79 = 2,806 (l)

c, n hh khí = 61,975/ 24,79 = 2,5 (mol)

   Đặt nO2 = x (mol)

    Dựa vào pt => NO2 = 4x

   => x + 4x = 2,5 <=> x = 0,5 (mol)

   Theo pt => nMg(NO3)2 = 2x = 1 (mol)

  nMg(NO3)2 cần dùng = 1/70 . 100 = 10/7 (mol)

 m Mg(NO3)2 = 10/7 .148 = 211,429 (g)