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The corrosion of aluminum in cookware is prevented as the aluminum metal reacts with oxygen gas in the air producing a protective coat of aluminum oxide. Aluminum) Al+3

(Oxygen) O-2

Al + O2 ------> Product(?
a) Write the balanced equation for the reaction.

b) How many grams of oxygen gas is used in the reaction if 3.25 moles of aluminum oxide is produced?

c) How many grams of aluminum oxide will be produced by the reaction

d) If the reaction gave an actual yield of 228.4 g, what is the percent yield for the reaction?

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Respuesta :

Answer:

Explanation:

 2Al    +    3 O₂   =    Al₂O₃

2 moles    3 moles        1 mole

b )

1 mole of aluminium oxide is produced from 3 moles of oxygen gas

3.25 moles of aluminium oxide is produced from 3 x 3.25 moles of oxygen

= 9.75 moles of oxygen

= 9.75 x 32grams of oxygen

= 312 grams of oxygen.

c )

3.25 moles of aluminium oxide = 3.25 x 102 = 331.5 grams of aluminium oxide

( molecular weight of aluminium oxide = 102 )

d ) actual yield = 228.4 gram

percent yield = 228.4 x 100 / 331.5

= 69 % approx .

adefioyelaoye
  • The balanced equation for the reaction is :

2 Al    +    3 O₂   =    Al₂O₃

  • Mass of oxygen gas if 3.25 moles of aluminum oxide is produced

3 moles of oxygen gas produces 1 mole of aluminium oxide

3.25 moles of aluminium oxide is produced from 3 x 3.25 moles of oxygen which is  9.75 moles of oxygen

Mass= mole × molar mass

= 9.75 x 32g of oxygen

= 312 g of oxygen.

  • The grams of aluminum oxide produced will be

Mass= mole x  molar mass

3.25 moles of aluminium oxide

= 3.25 x 102

= 331.5 g of aluminium oxide

  • Percent yield when Actual yield = 228.4g

Percent yield = 228.4 x 100 / 331.5

= 68.89 %

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