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Given the balanced equation representing a reaction: 4Al(s) + 3O2(g) → 2Al2O3(s) How many moles of Al(s) react completely with 4.50 moles of O2(g) to produce 3.00 moles of Al2O3(s)?

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We must consider the relation between Al and [tex]O_2[/tex]:

[tex]\frac{3\ mol\ O_2}{4\ mol\ Al} = \frac{4,5\ mol\ O_2}{x}\ \to\ x = \frac{4,5\ mol\ O_2\cdot 4\ mol\ Al}{3\ mol\ O_2} = \bf 6\ mol\ Al[/tex]

As you can see, the number of moles of Al is double to the number of moles of [tex]Al_2O_3[/tex], keeping constant the proportion of both substances.

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