a. Phản ứng xảy ra:
\[2Mg+O_2\to 2MgO\]
b. \[n_{Mg}=\dfrac{7,2}{24}=0,3(mol)\]
Theo PTHH: $n_{Mg}=n_{MgO}=0,3(mol)$
$\to m_{MgO}=0,3.40=12(g)$
c. \[n_{O_2}=\dfrac{1}{2}.n_{Mg}=\dfrac 12. 0,3=0,15(mol)\]
\[\to V_{O_2}=0,15.22,4=3,36(l)\]
\[\to V_{kk}=\dfrac{3,36}{20\%}=16,8(l)\]