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Đáp án + Giải thích các bước giải:

Câu 4:

$\rm 2Cu+O_2 \xrightarrow{t^o} 2CuO\\ 2Mg+O_2 \xrightarrow{t^o} 2MgO\\ n_{Cu}=\dfrac{6,4}{64}=0,1(mol)\\ n_{Mg}=\dfrac{4,8}{24}=0,2(mol)\\ \xrightarrow{Theo\ pt:} n_{O_2}=\dfrac{1}{2}n_{Cu}+\dfrac{1}{2}n_{Mg}=0,15(mol)\\ V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{0,15.22,4}{20\%}=16,8(l)$

Câu 5:

$\rm 4Al+3O_2 \xrightarrow{t^o} 2Al_2O_3\\ 3Fe+2O_2 \xrightarrow{t^o} Fe_3O_4\\ 2Zn+O_2 \xrightarrow{t^o} 2ZnO\\ n_{Al}=\dfrac{5,4}{27}=0,2(mol)\\ n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ n_{Zn}=\dfrac{19,5}{65}=0,3(mol)\\ \xrightarrow{Theo\ pt:} n_{O_2}=\dfrac{3}{4}n_{Al}+\dfrac{2}{3}n_{Fe}+\dfrac{1}{2}n_{Zn}=\dfrac{11}{30}mol)\\ V_{kk}=\dfrac{V_{O_2}}{20\%}=\dfrac{\dfrac{11}{30}.22,4}{20\%}=41,067(l)$

Câu 6:

$\rm a)\\ 4K+O_2 \xrightarrow{t^o} 2K_2O\\ n_{O_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ n_K=4n_{O_2}=0,6(mol)\\ m_K=0,6.39=23,4(g)\\ b)\\ n_{K_2O}=2n_{O_2}=0,3(mol)\\ m_{K_2O}=0,3.94=28,2(g)$