\(n_P=\dfrac{31}{31}=1\left(mol\right)\)
`4P+5O_2->(t^o)2P_2O_5`
4 5 2 ( mol )
1 0,5 ( mol )
\(m_{P_2O_5}=0,5.142=71\left(g\right)\)
$n_P = \dfrac{31}{31} = 1(mol)$
$4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$n_{P_2O_5} = \dfrac{1}{2}n_P = 0,5(mol)$
$m_{P_2O_5} = 0,5.142 = 71(gam)$
\(n_P=\dfrac{31}{31}=1\left(mol\right)\)
\(4P+5O_2\rightarrow2P_2O_5\)
1 → 0,5
\(\Rightarrow m_{P_2O_5}=0,5\left(31\cdot2+16\cdot5\right)=71\left(g\right)\)