Đáp án:

4P+5O2=2P2O5(1)

 x     1,25x

S+O2=SO2(2)

 y    y

nO2=0,45mol

31x+32y=12,6

1,25x+y=0,45

=>x=0,2

y=0,2

mS=0,2.32=6,4g

mP=0,2.31=6,2g

nO2(1)=1,25x=1,25.0,2=0,25mol

mO2(1)=0,25.32=8g

nO2(2)=0,2mol

mO2(2)=0,2.32=6,4g

$n_{O_2}=10,08/22,4=0,45mol$

Gọi $n_S=a(mol);n_P=b(mol)(a,b>0)$

$S+O_2\overset{t^o}{\longrightarrow}SO_2(1)$

$4P+5O_2\overset{t^o}{\longrightarrow}2P_2O_5(2)$

Ta có :

$m_{hh}=32a+31b=12,6$

$n_{O_2}=n_{O_2(1)}+n_{O_2(2)}=a+1,25b=0,45$

Ta có hpt :

$\left\{\begin{matrix} 32a+31b=12,6 & \\ a+1,25b=0,45& \end{matrix}\right.$

⇔$\left\{\begin{matrix} a=0,2 & \\ b=0,2& \end{matrix}\right.$

$⇒m_S=0,2.32=6,4g$

$m_{O_2(1)}=0,2.32=6,4g$

$m_P=0,2.31=6,2g$

$m_{O_2(2)}=5/4.0,2.32=8g.$