$\text{#Euphoria}$

$\text{a,Na$_{2}$O:}$

$\text{%m$_{Na}$=$\frac{23.2}{23.2+16}$.100% ≈74,19%}$

$\text{%m$_{O}$=(100-74,19)%=25,81%}$

$\text{Cu(OH)$_{2}$:}$

$\text{%m$_{Cu}$=$\frac{64}{64+17.2}$.100% ≈65,31%}$

$\text{%m$_{O}$=$\frac{2.16}{64+17.2}$.100% ≈32,65%}$

$\text{%m$_{H}$=(100-65,31-32,65)%=2,04%}$

$\text{K$_{2}$SO$_{4}$:}$

$\text{%m$_{K}$=$\frac{39.2}{174}$.100% ≈44,83%}$

$\text{%m$_{S}$=$\frac{32}{174}$.100% ≈18,39%}$

$\text{%m$_{O}$=(100-44,83-18,39)%=36,78%}$

$\text{MgCl$_{2}$:}$

$\text{%m$_{Mg}$=$\frac{24}{95}$.100% ≈25,26%}$

$\text{%m$_{Cl}$=(100-25,26)%=74,74%}$

$\text{HNO$_{3}$:}$

$\text{%m$_{H}$=$\frac{1}{63}$.100% ≈1,59%}$

$\text{%m$_{N}$=$\frac{14}{63}$.100% ≈22,22%}$

$\text{%m$_{O}$=(100-1,59-22,22)%=76,19%}$

$\text{Al$_{2}$(SO$_{4}$)$_{3}$:}$

$\text{%m$_{Al}$=$\frac{27.2}{342}$.100% ≈15,79%}$

$\text{%m$_{S}$=$\frac{3.32}{342}$.100% ≈28,07%}$

$\text{%m$_{O}$=(100-15,79-28,07)%=56,14%}$

$\text{b,Gọi CTHH của h/c khí đó là N$_{x}$H$_{y}$}$

$\text{h/c đó có tỉ khối với H$_{2}$ là 8,5}$

$\text{⇒M$_{NxHy}$=8,5.2=17(g/mol)}$

$\text{⇒x:y=$\frac{82,35}{14}$: $\frac{17,65}{1}$=1:3 }$

$\text{⇒x=1;y=3}$

$\text{⇒CTHH h.c khí đo là NH$_{3}$}$ 

Đáp án+Giải thích các bước giải:

`a)` 

`M_{Na_2O} = 23xx2 + 16 = 62` `$(g/mol)$

`%Na = {23.2.100}/{62}` `= 74,19%`

`%O = 100 - 74,19 = 25,81%`

`M_{Cu(OH)_2} = 64 + (16 + 1)xx2 = 98` $(g/mol)$

`%Cu = {64.100}/{98}` `= 65,3%`

`%O = {16.2.100}/{98}` `= 32,6%`

`%H = 100 - 65,3 - 32,6 = 2,1%`

`M_{K_2SO_4} = 39xx2 + 32 + 16xx4 = 174` $(g/mol)$

`%K = {39.2.100}/{174}` `= 44,8%`

`%S = {32.100}/{174}` `= 18,3%`

`%O = 100 - 44,8 - 18,3 = 36,9%`

`M_{MgCl_2} = 24 + 35,5xx7 = 95` $(g/mol)$

`%Mg = {24.100}/{95}` `= 25,26%`

`%Cl = 100 - 25,26 = 74,74%`

`M_{HNO_3} = 1 + 14 + 16xx3 = 63` $(g/mol)$

`%H = {1.100}/{63}` `= 1,5%`

`%N = {14.100}/{63}` `= 22,2%`

`%O = 100 - 1,5 - 22,2 = 76,3%`

`M_{Al_2(SO_4)_3} = 27xx2 + (32 + 16xx4)xx3 = 342` $(g/mol)$

`%Al = {27.2.100}/{342}` `= 15,7%`

`%S = {32.3.100}/{342}` `= 28%`

`%O = 100 - 15,7- 28 = 56,3%`

`b)` Ta có :

`d_{hc//H_2} = 8,5`

`=>` `M_{hc} = 8,5xxH_2 = 8,5xx1xx2 = 17` $(g/mol)$

`%H = 100 - 82,35 = 17,65%`

`CTHH` dạng chung : `N_xH_y`

`%N = {14x.100}/{17}` `= 82,35%`

`->` `x=1`

`%H = {1y.100}/{17}` `= 17,65%`

`->` `x=3`

Vậy `CTHH` là : `NH_3`