Mai intai aflam numarul de moli de fosfor (P):
[tex]\displaystyle\nu _P = \frac{m_P}{\mu_P}\\\\m_P=62\:g\\\mu_P=31\:g/mol\\\\\displaystyle\nu _P = \frac{62}{31}=2\:moli[/tex]
[tex]\underset{2\:moli}{2P}+\dfrac52O_2 \longrightarrow \underset{x\:moli}{P_2O_5}\\\\\dfrac22=\dfrac1x\implies 1=\dfrac1x\implies x=1\\\\\nu _{P_2O_5}=x=1 mol[/tex]
[tex]\mu _{P_2O_5}=2A_P+5A_O=2\cdot31+5\cdot16=62+80=142\:g/mol\\\\m_{P_2O_5}=\mu_{P_2O_5}\cdot\nu_{P_2O_5}=142\cdot1=142\:g[/tex]
Raspuns: 142 g de oxid de fosfor
