oblicz masę tlenku glinu jaki powstanie w wyniku reakcji 108g glinu z tlenem.
108u 204u
4Al + 3O₂ --> 2Al₂O₃
108g x
z proprocji wyliczam x
x = 204g
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Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
Mmol Al₂O₃ = 2·27+3·16=102g
108g x g
4Al + 3O₂ ⇒ 2Al₂O₃
27·4 102·2
108(102·2)=x(27·4)
22032=108x
x=204g
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108u 204u
4Al + 3O₂ --> 2Al₂O₃
108g x
z proprocji wyliczam x
x = 204g
----------------------------------------------------------------------------------------------------
Litterarum radices amarae sunt, fructus iucundiores
Pozdrawiam :)
Mmol Al₂O₃ = 2·27+3·16=102g
108g x g
4Al + 3O₂ ⇒ 2Al₂O₃
27·4 102·2
108(102·2)=x(27·4)
22032=108x
x=204g