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Trả lời:
Bài 1:
$Pt:\,4P+5O_2\xrightarrow{t^o}2P_2O_5$
$n_{O_2}=\dfrac{2,24}{22,4}=0,1\,(mol)$
$⇒n_P=\dfrac{4}{5}.n_{O_2}=0,08\,(mol)$
$⇒m_P=0,08.31=2,48\,(g).$
Bài 2:
$Pt:\,3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$
$n_{O_2}=2.n_{Fe_3O_4}=2.0,2=0,4\,(mol)$
$⇒V_{O_2}=0,4.22,4=8,96\,(l).$