Under certain conditions the rate of this reaction is zero order in ammonia with a rate constant of 0.0089 M. s^-1: 2NH3(g) → N2(g)+ 3H2(g)
Suppose a 5.0 L flask is charged under these conditions with 400. mmol of ammonia. How much is left 2.0 s later? You may assume no other reaction is important.
Be sure your answer has a unit symbol, if necessary, and round it to 2 significant digits.

Explanation:

The rate of reaction will not depend upon concentration of reactant . It will be always constant and equal to .0089M s⁻¹.

Initial moles of reactant = 400 x 10⁻³ mole in 5 L

molarity = 400 x 10⁻³ /5 M

= 80 x 10⁻³ M .

= .08M

no of moles reacted in 2 s  = .0089 x 2

= .0178 M

concentration left = .08 - .0178 M

= .0622 M .

No of moles left in 5 L

= 5 x .0622 = .31 moles .

the answer is 0.42 is that is an option. this question provides no info