Question #b6f49

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Question #b6f49

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Answer 1 · 2018-01-13T06:24:47

$164 g A l_{2} O_{3}$ $color(blue)(164gAl_2O_3)$

Explanation:

First, write a balanced equation:
$4 A l (s) + 3 O_{2} (g) \to 2 A l_{2} O_{3}$ $4Al(s) + 3O_2(g) -> 2Al_2O_3$

Then, use the equation to determine how many moles of $A l_{2} O_{3}$ $Al_2O_3$ will be created when 3.23 moles of aluminum reacts:
$3.23 m o l A l \cdot (\frac{2 m o l A l_{2} O_{3}}{4 m o l A l}) = 1.615 m o l A l_{2} O_{3}$ $3.23molAl*((2molAl_2O_3)/(4molAl)) = 1.615molAl_2O_3$

Finally, convert into grams:
$1.615 m o l A l_{2} O_{3} \cdot (\frac{102 g A l_{2} O_{3}}{1 m o l A l_{2} O_{3}}) = 164.73 \approx 164 g A l_{2} O_{3}$ $1.615molAl_2O_3*((102gAl_2O_3)/(1molAl_2O_3)) = 164.73 ~~ 164gAl_2O_3$

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Question #b6f49

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