Calculate the number of oxygen atoms required to combine with 7gm of n2
Answers
Answered by
11
Hello there,
● Complete question is -
Q. Calculate the number of oxygen atoms required to combine with 7gm of N2 to form N2O3.
● Answer -
N = 4.517×10^23 atoms
● Explaination -
No of moles of N2 is calculated by -
n = 7/28 = 0.25 moles
Reaction for production of N2O3 -
2N2 + 3O2 ---> 2N2O3
2 moles of N2 react with 3 moles of O2
∴ 0.25 moles of N2 react with 0.375 moles of O2.
Total no of oxygen atoms is thus -
N = n × Na × 2
N = 0.375 × 6.022×10^23 × 2
N = 4.517×10^23 atoms
Threfore, 4.5×10^23 O atoms are required for given reaction.
Hope this helps you...
● Complete question is -
Q. Calculate the number of oxygen atoms required to combine with 7gm of N2 to form N2O3.
● Answer -
N = 4.517×10^23 atoms
● Explaination -
No of moles of N2 is calculated by -
n = 7/28 = 0.25 moles
Reaction for production of N2O3 -
2N2 + 3O2 ---> 2N2O3
2 moles of N2 react with 3 moles of O2
∴ 0.25 moles of N2 react with 0.375 moles of O2.
Total no of oxygen atoms is thus -
N = n × Na × 2
N = 0.375 × 6.022×10^23 × 2
N = 4.517×10^23 atoms
Threfore, 4.5×10^23 O atoms are required for given reaction.
Hope this helps you...
Answered by
3
Explanation:
see the above image for answer
Attachments:
Similar questions