Question

the weight of oxygen required to completely react with 27 g of Al is​

Answer 1

4Al+3O2------> 2 Al2 O3

atomic mass of Al=27g

atomic mass of O2= 32g

mole ratio of Al and O2 can be given as -:

Al:O2

4:3

1:4/3

mass of Oxygen required =24g

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Answer 2

Answer:

Explanation:

Chemical Reactions and Stoichiometry:-

4Al+3O2------>2Al2O3

From the reaction,we find four moles of Aluminium react with three moles of Oxygen.

Now, from Avogadro's law,Weight of a mole of substance=Its gram molecular mass.

Molecular mass of Aluminium=27g

So we need to calculate weight of oxygen for one mole of Aluminium.

3 moles of oxygen--------4 moles of aluminium

=>(3*32)g oxygen--------(4*27)g aluminium

Hence,for 27g aluminium, weight of oxygen=(3*32)/4=24g oxygen.

So 24g oxygen is needed.