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Rama09 [41]
3 years ago
8

If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced

Chemistry
1 answer:
marissa [1.9K]3 years ago
8 0

The amount in grams of  Al₂O₃ produced is approximately 6.80 g.

Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:

AL + O₂ → Al₂O₃

Let's balance it

4AL + 3O₂ → 2Al₂O₃

4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,

Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,

Atomic mass of AL = 27 g

Molar mass of  Al₂O₃ = 101.96 g/mol

4(27 g) of AL gives 2(101.96 g) of  Al₂O₃

3.6 g of AL will give ?

cross multiply

mass of  Al₂O₃ produced = 3.6 × 203.92 / 108   = 734.112 / 108 = 6.797

mass of  Al₂O₃ produced = 6.80 g.

read more: brainly.com/question/23982245?referrer=searchResults

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Answer:

The answer to your question is 80.3%

Explanation:

Data

Percent by mass of F

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Process

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2.- Use proportions and cross multiplications to find the percent by mass of F. The molar mass of NF₃ is equal to 100%.

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The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.

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