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The chemical equation below shows the formation of aluminum oxide [tex]$\left( Al_2O_3 \right)$[/tex] from aluminum [tex]$\left( Al \right)$[/tex] and oxygen [tex]$\left( O_2 \right)$[/tex].

[tex]\[ 4 Al + 3 O_2 \rightarrow 2 Al_2O_3 \][/tex]

The molar mass of [tex]$O_2$[/tex] is [tex]$32.0 \, \text{g/mol}$[/tex]. What mass, in grams, of [tex]$O_2$[/tex] must react to form 3.80 mol of [tex]$Al_2O_3$[/tex]?

A. 60.8 grams
B. 81.1 grams
C. 122 grams
D. 182 grams

Sagot :

GinnyAnswer
To determine the mass of [tex]\( O_2 \)[/tex] needed to form 3.80 moles of [tex]\( Al_2O_3 \)[/tex], we can follow these detailed steps:

1. Understanding the Mole Ratio:
Based on the balanced chemical equation:
[tex]\[ 4 \, \text{Al} + 3 \, \text{O}_2 \rightarrow 2 \, \text{Al}_2\text{O}_3 \][/tex]
This equation tells us that 3 moles of [tex]\( O_2 \)[/tex] react to form 2 moles of [tex]\( Al_2O_3 \)[/tex].

2. Calculate the Moles of [tex]\( O_2 \)[/tex] Required:
We need to find out how many moles of [tex]\( O_2 \)[/tex] are required to produce 3.80 moles of [tex]\( Al_2O_3 \)[/tex]:
[tex]\[ \text{To produce } 2 \text{ moles of } Al_2O_3, \text{ we need } 3 \text{ moles of } O_2 \][/tex]
Using the stoichiometric coefficients to find moles of [tex]\( O_2 \)[/tex] needed:
[tex]\[ \text{Moles of } O_2 = \left( \frac{3}{2} \right) \times 3.80 = 5.70 \text{ moles of } O_2 \][/tex]

3. Determine the Mass of [tex]\( O_2 \)[/tex] Needed:
The molar mass of [tex]\( O_2 \)[/tex] is given as [tex]\( 32.0 \, \text{g/mol} \)[/tex]. To find the mass in grams, we multiply the moles of [tex]\( O_2 \)[/tex] needed by the molar mass:
[tex]\[ \text{Mass of } O_2 = 5.70 \, \text{moles} \times 32.0 \, \text{g/mol} = 182.4 \, \text{g} \][/tex]

Therefore, the mass of [tex]\( O_2 \)[/tex] required to react and form 3.80 moles of [tex]\( Al_2O_3 \)[/tex] is 182 grams.
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