How many grams of H2O2 were in the original sample?
Hydrogen peroxide (H₂O₂) decomposes in the presence of a catalyst, resulting in the formation of water and oxygen. A catalyst was added to a 5.60 mL sample of hydrogen peroxide solution at 20.0 °C, leading to the collection of 49.5 mL of gas over water, while the total pressure measured was 764.4 mmHg. To determine the amount of hydrogen peroxide in grams present in the original sample, please reference the vapor pressure of water at these conditions from a reliable table.
2 Answers
WVP = 17.5 mm Hg
Gas pressure = 764.4 – 17.5 = 746.9 mm Hg
The temperature of the hydrogen peroxide solution must be converted from ˚C to ˚K by adding 273.
T = 20 + 273 = 293
Let the volume be converted to liters.
V = 0.0495 liter
Let’s use the following equation to equation to determine the volume of oxygen at standard temperature and pressure.
P1 V1 ÷ T1 = P2 V2 ÷ T2
746.9 0.0495 ÷ 293 = 760 V2 ÷ 273
V2 = 10,093.213315 ÷ 222,680
The volume is approximately 0.0453 liter. The volume of one mole of gas at standard temperature and pressure is 22.4 liters.
n = 10,093.213315 ÷ 222,680) ÷ 22.4
This is approximately 0.00202 mole of oxygen.
H2O2 → H2O + O2
To balance this equation, make 2 H2O2 and 2 H2O.
2 H2O2 → 2 H2O + O2
According to the coefficients in the balanced equation, two moles of hydrogen peroxide will produce two moles of water and one mole of oxygen. This means the number of moles hydrogen peroxide is twice the number of moles of oxygen.
n = (10,093.213315 ÷ 222,680) ÷ 11.2
This is approximately 0.00404 mole of hydrogen peroxide.
The mass of one mole of hydrogen peroxide is 34 grams.
This is approximately 0.138 grams. This is the only way that I know how to solve this type of problem. I hope this is helpful for you.
Grams of H2O2.....
2H2O2(l) --> 2H2O(l) + O2(g)
...?g .............................. 49.5 mL, 20C, 764.4 Torr
Find the moles of O2 ............ use Dalton's law and the ideal gas equation
Double that to get moles of H2O2 ....... use the unit factor method for steps 2 and 3
Calculate mass of H2O2.
Vapor pressure of water at 20.0C is 17.5 Torr. Subtract to get the pressure of "dry" O2.
PV = nRT ................ ideal gas equation
n = PV / (RT) .......... solve for n (moles)
n = 746.9 Torr x 0.0495L / 62.36 LTorr/molK / 293.15K
n = 0.00202 mol O2
0.00202 mol O2 x (2 mol H2O2 / 1 mol O2) x (34.0g H2O2 / 1 mol H2O2) = 0.138g H2O2
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