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Mars2501 [29]
3 years ago
10

How many moles of neon occupy a volume of 14.3 l at stp? how many moles of neon occupy a volume of 14.3 l at stp? 1.57 moles 0.6

38 moles 36.7 moles 32.0 moles 6.45 moles?
Chemistry
1 answer:
andriy [413]3 years ago
5 0
The  number   of  neon  moles  that  occupy  a volume of 14.3 l  at STP is calculated as follows

At STP 1  mole = 22.4 liters

what about  14.3 liters

by cross  multiplication
= (1 mole x 14.3 l)/22.4 l =0.638  moles  of  neon


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48g of 02 produce how many grams of Al2O3
bixtya [17]

Taking into account the reaction stoichiometry,  102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

4 Al + 3 O₂  → 2 Al₂O₃

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Al: 4 moles
  • O₂: 3 moles
  • Al₂O₃: 2 moles

The molar mass of the compounds is:

  • Al: 27 g/mole
  • O₂: 32 g/mole
  • Al₂O₃: 102 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Al: 4 moles ×27 g/mole= 108 grams
  • O₂: 3 moles ×32 g/mole= 96 grams
  • Al₂O₃: 2 moles ×102 g/mole= 204 grams

<h3>Mass of Al₂O₃ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 96 grams of O₂ form 204 grams of Al₂O₃, 48 grams of O₂ form how much mass of Al₂O₃?

mass of Al_{2} O_{3} =\frac{48 grams of O_{2} x204 grams of Al_{2} O_{3}}{96 grams of O_{2}}

<u><em>mass of Al₂O₃= 102 grams</em></u>

Finally, 102 grams of Al₂O₃ are formed when 48 grams of O₂ react.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

3 0
2 years ago
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