Which smallest whole-number coefficients balance N2 + O2 giving dinitrogen pentoxide, N2O5?

1. Start from N2 + O2 giving N2O5, which has 2 O atoms on the left and 5 on the right. 2. The lowest common multiple of 2 and 5 is 10, so set O2 to 5 and N2O5 to 2, giving 10 O atoms each side. 3. That fixes oxygen but leaves 2 N atoms on the left and 4 on the right. 4. Restore nitrogen by setting the N2 coefficient to 2, giving 4 N atoms on each side. 5. Final check: N is 4 and 4, O is 10 and 10, so 2 N2 + 5 O2 giving 2 N2O5 is balanced. 6. Option A leaves nitrogen unbalanced at 2 against 4, and option C leaves oxygen at 6 against 5. 7. Option D is a valid ratio only if oxygen is doubled too, so it is not balanced as written. _Source: OpenStax Chemistry (CC BY 4.0), Ch 4 "Stoichiometry of Chemical Reactions", section 4.1 Writing and Balancing Chemical Equations_