Acidul fosforic: H3PO4
[tex]M_{H3PO4}=3*1+31+4*16=98 \ \ \ =\ \textgreater \ \ \ u_{H3PO4}=98 \ g/mol[/tex]
(Am notat cu M masa moleculara si cu u masa molara)
Aplicam regula de 3 simpla:
98 g H3PO4................4*16 g O
m..........................640 g O
[tex]m= \dfrac{640*98}{4*16}=980\ g\ H3PO4 [/tex]
In moli:
[tex]n_{H3PO4}= \dfrac{980}{98}=10 \ moli [/tex]
