Đáp án:

\[a)\, m_{P\, \text{dư}}=1,24\text{(g)}\\ b)\, m_{P_2O_5}=11,36\text{(g)}\] 

Giải thích các bước giải:

\(a)\, 4P+5O_2\xrightarrow{t^{\circ}} 2P_2O_5\\ n_P=\dfrac{6,2}{31}=0,2\text{(mol)}\\ n_{O_2}=\dfrac{4,48}{22,4}=0,2\text{(mol)}\)

\(\dfrac{0,2}4>\dfrac{0,2}5\to\)Sau phản ứng $P$ dư, $O_2$ hết

\(\to n_{P\, \text{phản ứng}}=\dfrac{0,2\times 4}5=0,16\text{(mol)}\to n_{P\, \text{dư}}=0,2-0,16=0,04\text{(mol)}\to m_{P\, \text{dư}}=31\times 0,04=1,24\text{(g)}\)

\(b)\, n_{P_2O_5}=\dfrac{0,2\times 2}{5}=0,08\text{(mol)}\to m_{P_2O_5}=142\times 0,08=11,36\text{(g)}\)

a ) $n_{P}= $6,2/31$ = $0,2 mol$

$n_{O_2}$ = $4,48/22,4$ = $0,2 mol$

PTHH : $4P$ + $5O_2$ → $2P_2O_5$

→ $n_{O_P}$ $dư$ $ 0,01 mol$

$m_{O_P}$= $0,01 × 31 $ = $0,31g$

b) $n_{P_2O_5}$= $0,08 mol$

$m_{P_2O_5}$ = $0,08 ×142 = 11,36 g$