Đáp án+ Giải thích các bước giải:

 a) $n_{Ba} = n_{Ba(NO_3)_2}$ $=$ $1,5$ (mol) 

$n_{N} = 2n_{Ba(NO_3)_2}$ $=$ $2. 1,5$ $=$ $3$ (mol) 

$n_{O} = 6n_{Ba(NO_3)_2}$ $= 6. 1,5$ $=$ $9$ (mol)

b) $n_{Ba} = n_{Ba(NO_3)_2}$ $=$ $0,5$ (mol) 

`->` $m_{Ba} = 0,5. 137 = 68,5$ (g) 

$n_{N} = 2n_{Ba(NO_3)_2}$ $=$ $2. 0,5$ $=$ $1$ (mol) 

`->` $m_{N} = 1. 14 = 14$ (g) 

$n_{O} = 6n_{Ba(NO_3)_2}$ $= 6. 0,5$ $=$ $3$ (mol)

`->` $m_{O} = 3. 16 = 48$ (g) 

c) $n_{Ba} =$ `(2,74)/137 = 0,02` (mol) 

`->` $n_{Ba(NO_3)_2} = n_{Ba} = 0,02$ (mol) 

`->` $m_{Ba(NO_3)_2} = 0,02. 261$ $=$ $5,22$ (g) 

d) `%Ba = 137/261 .100 ≈ 52,49%`

`%N = 28/261 .100 ≈ 10,73%` 

`%O = 100% - 52,49% - 10,73% = 36,78%`

`#Ly`

Ta có: `Ba(NO_3)_2` có thể viết thành: `BaN_2O_6`

$a)$

Trong $1,5$ $mol$ `BaN_2O_6`, có:

`n_(Ba` `=` `1,5.1=1,5(mol)`

`n_(N` `=` `1,5.2=3(mol)`

`n_(O` `=` `1,5.6=9(mol)`

$b)$

Trong $0,5$ $mol$ `BaN_2O_6`, có:

`n_(Ba` `=` `0,5.1=0,5(mol)`

`->` `m_(Ba` `=` `0,5.137=68,5(g)`

`n_(N` `=` `0,5.2=1(mol)`

`->` `N=1.14=14(g)`

`n_(O` `=` `0,5.6=3(mol)`

`->` `n_(O` `=` `3.16=48(g)`

$c)$

`n_(Ba` `=` `(2,74)/(137)=0,02(mol)`

Mà `n_(Ba` `=` `n_(BaN_2O_6`

`=>` `n_(BaN_2O_6` `=0,02(mol)`

`m_(Ba(NO_3)_2` `=` `0,02.261=5,22(g)`

$d)$

`%Ba` `=` `(137)/(261).100%≈52,49%`

`%N` `=` `(2.14)/(261).100%≈10,73%`

`%O` `=` `100-52,49-10,73=36,78%`