Phương pháp giải:

Lời giải chi tiết:

a/ HCl                         

 \({M_{HCl}} =1 + 35,5 = 36,5(g)\)

 \( => \% H = \frac{{{m_H}}}{{{M_{HCl}}}}.100\%  = \frac{1}{{36,5}}.100\%  = 2,74\% \)

%Cl = 100% - % H = 97,26%)

b/ MgSO₄                      

 \( => \% Mg = \frac{{{m_{Mg}}}}{{{M_{MgS{O_4}}}}}.100\%  = \frac{{24}}{{120}}.100\%  = 20\% \)

\(\% O = \frac{{{m_O}}}{{{M_{MgS{O_4}}}}}.100\%  = \frac{{16.4}}{{120}}.100\%  = 53,33\% \)

  %S = 100% - % Mg - %O = 26,67%

c/ FeS₂                                

 \({M_{Fe{S_2}}} = 56 + 32.2 = 120(g)\)

 \( => \% Fe = \frac{{{m_{Fe}}}}{{{M_{Fe{S_2}}}}}.100\%  = \frac{{56}}{{120}}.100\%  =46,67\% \)

%S = 100% - % Fe  = 53,33%

d/ CO₂

 \({M_{C{O_2}}} = 12 + 16.2 = 44(g)\)

 \( => \% C = \frac{{{m_C}}}{{{M_{C{O_2}}}}}.100\%  = \frac{{12}}{{44}}.100\%  = 27,27\% \)

 %O = 100% - %C = 72,73%