Đáp án + Giải thích các bước giải:

Bài 3:

$a)\%m_{Al}=\frac{27.2}{27.2+16.3}.100\%\approx 52,94\%$

$\%m_O=100-52,94=47,06\%$

$b)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$n_{Al_2O_3}=\frac{20,4}{102}=0,2(mol)$

Theo PT: $n_{O_2}=\frac{3}{2}n_{Al_2O_3}=0,3(mol)$

$\to m_{O_2}=0,3.32=9,6(g)$

$c)2KClO_3\xrightarrow{t^o}2KCl+3O_2$

Theo PT: $n_{KClO_3}=\frac{2}{3}n_{O_2}=0,2(mol)$

$\to m_{KClO_3}=0,2.122,5=24,5(g)$

Bài 4:

$a)n_{CuO}=\frac{48}{80}=0,6(mol)$

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

Theo PT: $n_{Cu}=n_{CuO}=0,6(mol)$

$\to m_{Cu}=0,6.64=38,4(g)$

$b)$ Theo PT: $n_{H_2}=n_{Cu}=0,6(mol)$

$\to V_{H_2(đktc)}=0,6.22,4=13,44(l)$

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