How many moles of Al2O3 will be formed when 27 grams of Al reacts completely with O2?

Explanation:

mole = mass in g/ molecular weight

molecular weight of Al2O3 =2(13) + 3(16) =74g/mol

reacting mass = 27 g

mole = 27/74 = 0.36mol

the answer would be a

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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