The number of oxygen atoms required to combine with 14 gm of nitrogen to form N2O3( 2 and 3 are - Science - Atoms and Molecules

Question

The number of oxygen atoms required to combine with 14 gm of
nitrogen to form N2O3( 2 and 3 are in subscript) when 80% of
nitrogen is converted to N203( 2 and 3 are in subscript)

Kirti Gogia · Accepted Answer

A balanced chemical equation showing the reaction between
N2 and O2 leading to the formation of
N2O3 is written as
2N2 + 3O2 â†’
2N2O3
We are given that 14 gram of nitrogen is present in the system,
and we have to calculate the number of oxygen atoms when 80 % of
this amount of N2 is converted to
N2O3 According to the above equation, 56 g of
N2 reacts with 96 g of O2 80 % of 14 gm is 11
2 gm Thus
56 g of N2 â†’ 96 g of
O2
So 1 gram of N2 â†’ ( 96 / 56) g of
O2
Hence 11 2 grams of N2 â†’ ( 96 X
11 2 )/ 56 g of O2
= 19 2 g of O2
Now the molar mass of O2 is 32 g / mol So
32 g of O2 = 1 mol = 6 023 X 1023
molecules of O2
Therefore 1 g of O2 = (6 023 X 1023) / 32
molecules of O2
Hence 19 2 g of O2 = (6 023 X 1023 X 19 2
) / 32 molecules of O2
= 3 614 X 1023 molecules of O2
Now 1 molecule of O2 = 2 atoms of oxygen
Therefore 3 614 X 1023 molecules of O2 = 2
X 3 614 X 1023 molecules of O2
= 7 228 X 1023 atoms of oxygen

The number of oxygen atoms required to combine with 14 gm of nitrogen to form N2O3( 2 and 3 are in subscript) when 80% of nitrogen is converted to N203( 2 and 3 are in subscript)