Phương pháp giải:

Lời giải chi tiết:

a.

\(S{O_2}:\% {m_{SO2}} = {{32} \over {32 + 2.16}}.100\%  = 50\% \)

%O = 100%-50%=50 %

\(S{O_3}:{\rm{ }}\% {\rm{ }}{m_{SO3}} = {{32} \over {32 + 16.3}}.100\%  = 40\% \)

% O= 100% −40%= 60%

b. 

\(KN{O_2}:\% K = {{39} \over {39 + 14 + 16.2}}.100\%  = 45,88\% \)

\(\% N = {{14} \over {39 + 14 + 16.3}}.100\%  = 13,86\% \)

%O= 100% −45,88%−16,47%=37,65%

\(KN{O_3}:{\rm{ }}\% {\rm{ }}K = {{39} \over {39 + 14 + 16.3}}.100\%  = 38,61\% \)

\(\% N{\rm{ }} = {{14} \over {39 + 14 + 16.3}}.100\%  = 13,86\% \)

% O=52,47% 

c.

\(Fe{\left( {OH} \right)_2}:{\rm{ }}\% {\rm{ }}Fe = {{56} \over {56 + 16.2 + 1.2}}.100\%  = 62,22\% \)

\(\% O = {{16.2} \over {56 + 16.2 + 1.2}}.100\%  = 35,55\% \)

%H=2,23% 

\(Fe{\left( {OH} \right)_3}:{\rm{ }}\% Fe{\rm{ }} = {{56} \over {56 + 16.3 + 1.3}}.100\%  = 52,34\% \)

\(\% O = {{16.3} \over {56 + 16.3 + 1.3}}.100\%  = 44,86\% \)

%H=2,8%