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Lynna [10]
3 years ago
5

How many grams of Aluminum Oxide are produced when 3.64 g of Aluminum reacts with an excess of Oxygen?

Chemistry
2 answers:
Vladimir79 [104]3 years ago
6 0

Answer:

<h2>6.877649 or <u>6.88 grams of Aluminum Oxide</u></h2>

Explanation:

- If there are 4 moles of Al that equals 107.926156 grams of 4Al.

- If there is 2 moles of Al2O3 that equals 203.9226 grams.

So...

If 107.926156 grams of 4Al equals 203.9226 grams of 2Al2O3 then, 3.64g of Al equals x.

fgiga [73]3 years ago
4 0
55664 grams of aluminum oxide gag
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El fluoruro de hidrógeno HF que se utiliza en
Blizzard [7]

Answer:

25.6g de HF son producidos

Explanation:

<em>...¿Cuánto HF es producido?</em>

Para resolver este problema debemos convertir la masa de cada reactivo a moles usando su masa molar. Como la reacción es 1:1, el reactivo con menor número de moles es el reactivo limitante. Con las moles del reactivo limitante podemos obtener las moles de HF y su masa así:

<em>Moles CaF2:</em>

Masa molar:

1Ca = 40g/mol

2F = 19*2 = 38g/mol

40+38 = 78g/mol

50g CaF2 * (1mol/78g) = 0.641 moles CaF2

<em>Moles H2SO4:</em>

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98g/mol

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Como las moles de CaF2 < Moles H2SO4: CaF2 es reactivo limitante.

<em>Moles HF usando la reacción:</em>

0.641 moles CaF2 * (2mol HF / 1mol CaF2) = 1.282 moles HF

<em>Masa HF:</em>

Masa molar:

1g/mol + 19g/mol = 20g/mol

1.282 moles HF * (20g/mol) =

<h3>25.6g de HF son producidos</h3>
8 0
3 years ago
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