In: Chemistry
While common consumer peroxide products fall in the 1-3% range, commercial solutions of peroxide can be as high as 30% peroxide. Could this method be used to measure peroxide concentrations in this range?
Problem 1 – Calculate the volume of gas you would expect to produce using 1.00 mL of 30% peroxide. Assume that the atmospheric pressure is 760 torr and the temperature of your water is 22°C. Can this volume be measured using the 50 mL buret in this apparatus?
Problem 2 – Will the peroxide still be the limiting reagent or will we need to increase the amount of permanganate in the reaction?
We have a 30% peroxide solution, i.e, 100 mL solution contains 30 g peroxide (density of water is assumed to be 1 g/mL).
Molar mass of hydrogen peroxide, H2O2 = (2*1.008 + 2*15.9994) g/mol = 34.0148 g/mol.
Mole(s) of peroxide corresponding to 30 g peroxide = (30 g)/(34.0148 g/mol) = 0.882 mole.
We have 100 mL solution; hence, the molar concentration is (0.882 mole)/[(100 mL)*(1 L/1000 mL)] = 8.82 mole/L.
We take 1.00 mL peroxide solution; hence, the number of moles of peroxide taken = (1.00 mL)*(1 L/1000 mL)*(8.82 mole/L) = 0.00882 mole.
Consider the decomposition of peroxide as below.
2 H2O2 (aq) ------> 2 H2O (l) + O2 (g)
Moles O2 produced = (0.00882 mole H2O2)*(1 mole O2/2 mole H2O2) = 0.00441 mole O2.
We have P = 760 torr = 1 atm and T = 22°C = (22 + 273) K = 295 K.
Use the ideal gas law.
P*V = n*R*T
====> (1.00 atm)*V = (0.00441 mole)*(0.082 L-atm/mol.K)*(295 K)
====> (1.00 atm)*V = 0.10776 L-atm
====> V = (0.10776 L-atm)/(1.00 atm) = 0.10776 L = (0.10776 L)*(1000 mL/1 L) = 107.76 mL
The volume of O2 gas produced is 107.76 mL. The said gas cannot be contained in a 50 mL buret (ans).
I need to know the experimental data to answer the remaining questions.