If 1/2 text{ mole of oxygen reacts with alumi | vedclass

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(B) The balanced chemical equation for the reaction is: $4Al + 3O_2 	o 2Al_2O_3$. From the stoichiometry,$3 	ext{ moles of } O_2 	ext{ react with }

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) The balanced chemical equation for the reaction is: $4Al + 3O_2 	o 2Al_2O_3$. From the stoichiometry,$3 	ext{ moles of } O_2 	ext{ react with } 4 	ext{ moles of } Al$. Therefore,$1 	ext{ mole of } O_2 	ext{ reacts with } 4/3 	ext{ moles of } Al$. For $1/2 	ext{ mole of } O_2$,the moles of $Al$ required are: $(4/3) 	imes (1/2) = 2/3 	ext{ moles}$. Mass of $Al = 	ext{moles} 	imes 	ext{molar mass} = (2/3) 	imes 27 	ext{ g} = 18 	ext{ g}$.

If $1/2 \text{ mole of oxygen reacts with aluminum to form } Al_2O_3, \text{ the mass of aluminum metal consumed in the reaction is } ..... \text{ g. } (Al=27)$

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