Đáp án:
\[n_{Cu}=\dfrac{38,4}{64}=0,6(mol)\]
a. $2Cu+O_2\to 2CuO$
b. $n_{O_2}=\dfrac{1}{2}\times n_{Cu}=0,3(mol)$
$\to V_{O_2}=0,3\times 22,4=6,72(l)$
c. \[2KClO_3\to 2KCl+3O_2\]
\[n_{KClO_3}=\dfrac{2}{3}\times n_{O_2}=\dfrac{2}{3}.0,3=0,2(mol)\]
\[\to m_{KClO_3}=0,2\times 122,5=24,5(g)\]