Đáp án + Giải thích các bước giải:

 1)

$M_{FeSO_4}=56 + 32 + 16.4= 152 (g/mol)$

`\text{%}m_{Fe}=(56)/(152).100\text{%}= 36,84\text{%}`

`\text{%}m_{S}=(32)/(152).100\text{%}= 21,05\text{%}`

`\text{%}m_{O}=100 - (36,84 + 21,05)= 42,11\text{%}`

2)

$M_{MgCl_2}=24 + 35,5.2= 95 (g/mol)$

`\text{%}m_{Mg}=(24)/(95) xx 100\text{%}= 25,26\text{%}`

`\text{%}m_{Cl}=100 - 25,26= 74,74\text{%}`

3)

$M_{Ca(NO_3)_2}=40 + (14 + 16.3).2= 164 (g/mol)$

`\text{%}m_{Ca}=(40)/(164).100\text{%}= 24,39\text{%}`

`\text{%}m_{N}=(14 xx 2)/(164).100\text{%}= 17,07\text{%}`

`\text{%}m_{O}=100 - (24,39 + 17,07)= 58,54\text{%}`

4)

$M_{Ag_2SO_4}=108.2 + 32 + 16.4= 312(g/mol)$

`\text{%}m_{Ag}=(108 xx 2)/(312).100\text{%}= 32,69\text{%}`

`\text{%}m_{S}=(32)/(312).100\text{%}= 10,26\text{%}`

`\text{%}m_{O}=100 - (32,69 + 10,26)= 57,05\text{%}`

5)

$M_{Ag_2SO_4}=108.3 + 31 + 16.4= 419 (g/mol)$

`\text{%}m_{Ag}=(108 xx 3)/(419).100\text{%}= 77,33\text{%}`

`\text{%}m_{S}=(32)/(419).100\text{%}= 7,64\text{%}`

`\text{%}m_{O}=100 - (77,33 + 7,64)= 15,36\text{%}`

6)

$M_{(C_2H_5COO)_2Ba}=(12.2 + 1.5 + 12 + 16 + 16).2 + 137= 283 (g/mol)$

`\text{%}m_{C}=(12 xx 6)/(283).100\text{%}= 25,44\text{%}`

`\text{%}m_{H}=(1 xx 10)/(283).100\text{%}= 3,53\text{%}`

`\text{%}m_{O}=(16 xx 4)/(283).100\text{%}= 22,61\text{%}`

`\text{%}m_{Ba}=100 - (25,44 + 3,53 + 22,61)= 48,42\text{%}`

7)

$M_{Ca(HSO_3)_2}=40 + (1 + 32 + 16.3).2= 202 (g/mol)$

`\text{%}m_{Ca}=(40)/(202).100\text{%}= 19,8\text{%}`

`\text{%}m_{H}=(1 xx 2)/(202).100\text{%}= 0,99\text{%}`

`\text{%}m_{S}=(32 xx 2)/(202).100\text{%}= 31,68\text{%}`

`\text{%}m_{O}=100 - (19,8 + 0,99 + 15,84)= 63,37\text{%}`

Đáp án:

Bạn tham khảo lời giải ở dưới nhé!!!

Giải thích các bước giải:

\(\begin{array}{l}
1,FeS{O_4}\\
\% Fe = \dfrac{{56}}{{56 + 32 + 16 \times 4}} \times 100\%  = 36,84\% \\
\% S = \dfrac{{32}}{{56 + 32 + 16 \times 4}} \times 100\%  = 21,05\% \\
\% O = 100\%  - 36,84\%  - 21,05\%  = 42,11\% \\
2,MgC{l_2}\\
\% Mg = \dfrac{{24}}{{24 + 35,5 \times 2}} \times 100\%  = 25,26\% \\
\% Cl = 100\%  - 25,26\%  = 74,74\% \\
3,Ca{(N{O_3})_2}\\
\% Ca = \dfrac{{40}}{{40 + 14 \times 2 + 16 \times 6}} \times 100\%  = 24,39\% \\
\% N = \dfrac{{14 \times 2}}{{40 + 14 \times 2 + 16 \times 6}} \times 100\%  = 17,07\% \\
\% O = 100\%  - 24,39\%  - 17,07\%  = 58,54\% \\
4,A{g_2}S{O_4}\\
\% Ag = \dfrac{{108 \times 2}}{{108 \times 2 + 32 + 16 \times 4}} \times 100\%  = 69,23\% \\
\% S = \dfrac{{32}}{{108 \times 2 + 32 + 16 \times 4}} \times 100\%  = 10,26\% \\
\% O = 100\%  - 69,23\%  - 10,26\%  = 20,51\% \\
5,A{g_3}P{O_4}\\
\% Ag = \dfrac{{108 \times 3}}{{108 \times 3 + 31 + 16 \times 4}} \times 100\%  = 77,33\% \\
\% P = \dfrac{{31}}{{108 \times 3 + 31 + 16 \times 4}} \times 100\%  = 7,4\% \\
\% O = 100\%  - 77,33\%  - 7,4\%  = 15,27\% \\
6,{({C_2}{H_5}COO)_2}Ba\\
\% C = \dfrac{{12 \times 6}}{{73 \times 2 + 137}} \times 100\%  = 25,44\% \\
\% H = \dfrac{{1 \times 10}}{{73 \times 2 + 137}} \times 100\%  = 3,53\% \\
\% O = \dfrac{{16 \times 4}}{{73 \times 2 + 137}} \times 100\%  = 22,61\% \\
\% Ba = 100\%  - 25,44\%  - 3,53\%  - 22,61\%  = 48,42\% 
\end{array}\)

\(\begin{array}{l}
7,Ca{(HS{O_3})_2}\\
\% Ca = \dfrac{{40}}{{40 + 2 + 32 \times 2 + 16 \times 6}} \times 100\%  = 19,8\% \\
\% H = \dfrac{2}{{40 + 2 + 32 \times 2 + 16 \times 6}} \times 100\%  = 0,99\% \\
\% S = \dfrac{{32 \times 2}}{{40 + 2 + 32 \times 2 + 16 \times 6}} \times 100\%  = 31,68\% \\
\% O = 100\%  - 19,8\%  - 0,99\%  - 31,68\%  = 47,53\% 
\end{array}\)