Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0067·Ms−1: 2N2O(g)→2N2(g)+O2(g)

Suppose a 4.0L ⁢ flask is charged under these conditions with 300.mmol of dinitrogen monoxide. After how much time is there only 150.mmol left? 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.

5.6 seconds

Explanation:

The reaction follows a zero-order in dinitrogen monoxide

Rate = k[N20]^0 = change in concentration/time

[N20]^0 = 1

Time = change in concentration of N2O/k

Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol

Initial concentration = moles/volume = 0.3/4 = 0.075

Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol

Concentration after t seconds = 0.15/4 = 0.0375 M

Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M

k = 0.0067 M/s

Time = 0.0375/0.0067 = 5.6 s

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%