For the first-order reaction, 2 n2o(g) → 2 n2(g) + o2(g), what is the concentration of n2o after 3 half-lives if 0.25 mol of n2o is initially placed into a 1.00-l reaction vessel? for the first-order reaction, 2 n2o(g) → 2 n2(g) + o2(g), what is the concentration of n2o after 3 half-lives if 0.25 mol of n2o is initially placed into a 1.00-l reaction vessel? 3.1 × 10-2 m 1.2 × 10-1 m 6.2 × 10-2 m 1.6 × 10-2 m

According to that each half-life we lose half of the concentration of N2O so,

Because we start with a concentration of N2O = 0.25 mol 
so after one half-life the concentration of N2O decrease to the half 0.25/2
= 12.5 x 10^-2 M
and after two half-lives the concentration of N2O of the one half-life decrease to the half (12.5 x 10^-2) / 2 
=6.25 x 10^-2 M
and after three half-lives the concentration of N2O of the two half-lives decrease to the half (6.25x10^-2) / 2
= 3.1 x 10^-2 M
∴ your correct answer is 3.1 x 10^-2 M

Fine out your self ponk