Question

Table 2: Temperature, Pressure, and Volume Data Temperature of Tap Water (°C) Room (or regional) Pressure (atm) Initial Volume of Air (mL) Final Volume of Air (after reaction) Volume of O2 Collected (Final Volume - Initial Volume) 32 (mL) 15 .99 35 67 Table 3: Reaction Time Data Total Reaction Time Time Reaction Started 10:10 Time Reaction Ended 10:10 47.51 (s)

Using the Ideal Gas Law (PV = nRT), calculate the grams of O₂ produced in the reaction. (Hint: solve for n, and then convert moles to grams. Don’t forget to convert your temperature from Celsius to Kelvin.) Show your work.

I know that Kelvin is 288 but beyond that, I'm not sure how to proceed Thank you!

Answer 1

Temperature = 15°c - 288k Pressure - 0.99 atm volume of O2 collected - 67 mL = 67x103 L Value of R in litre, adm unit is 0.082 L.atm.mollet Now, from, pv=MRT equation, PV n = RT 0.99 x 67 x103 0.082 x 288 - 2.8x103 mol molecular weight of O2 = 32 g/mol 2.8x10 3 mol O2 = 322.8x103 gram = 0.0899 groom. ..0.08 99 groam on will be produced in the reaction.