Question

missed this? watch kcv reaction stoichiometry, iwe stoichiometry. we stoichiometry: read section 4.3. you can click on the review link to access the section in e - text. balance the equation and calculate how many moles of o₂ form when each quantity of reactant completely reacts. n₂o₅(g)→no₂(g)+o₂(g) part a balance the equation. αn₂o₅(g)→βno₂(g)+γo₂(g) give your answer as an ordered set of numbers α, β, and γ. use the lowest ratio of integers for the coefficients. view available hint(s) α,β,γ= previous answers submit incorrect; try again part b complete previous part(s) part c complete previous part(s)

Explanation:

Step1: Balance nitrogen atoms

On the left - hand side, there are 2 nitrogen atoms in \(N_2O_5\). On the right - hand side, nitrogen is in \(NO_2\). To balance nitrogen, if we assume the coefficient of \(N_2O_5\) is \(\alpha\) and of \(NO_2\) is \(\beta\), for 2 nitrogen atoms in \(N_2O_5\), we need \(\beta = 2\alpha\). Let \(\alpha=2\), then \(\beta = 4\).

Step2: Balance oxygen atoms

The number of oxygen atoms in \(2N_2O_5\) is \(2\times5 = 10\). The number of oxygen atoms in \(4NO_2\) is \(4\times2=8\). The remaining oxygen atoms form \(O_2\). The number of oxygen atoms in \(O_2\) is \(10 - 8=2\), so the coefficient of \(O_2\) (\(\gamma\)) is 1.

Answer:

\(\alpha = 2,\beta = 4,\gamma = 1\)