A student performed the experiment as described in the lab manual, using 5.00 mL

Question

A student performed the experiment as described in the lab manual, using 5.00 mL of an aqueous hydrogen peroxide solution, with a density of 1.01 g/mL. The water temperature was 294 K and the barometric pressure was 31.50 in. Hg. After the student immersed the yeast in the peroxide solution, they observed a 43.34 mL change in system volume.

The barometric pressure, in torr, is ___________ torr. (1 dec places)

The water vapor pressure at the water temperature is __________ torr. (1 dec place)

The pressure exerted by the collected oxygen gas at the water temperature is _______ torr (1 dec place) or _________ atm (2 dec places).

The volume of collected oxygen gas is__________ L (4 significant figures).

Using R = 0.08206, the number of moles of collected oxygen gas is __________ (2 dec places) x 10________ (integer) mol.

So, based upon the number of moles of oxygen gas collected, the number of moles of peroxide in the original 5.00 mL sample is __________ (2 dec places) x 10__________(integer) moles.

So, the mass of peroxide (MW = 34.02), is__________ g (3 dec places).

Explanation / Answer

The reaction is 2H2O2----> 2H2O+O2 (g)

1 inch of mercury= 25.4 torr

31.5 inches of mercury= 800 torr

The water vapor pressure at 294 K is 18.65 torr

partial pressure of oxygen =Barometic pressure -vapor pressure of water vapor = 800- 18.65 torr = 781.35 torr =781.35 atm

Volume of oxygen collected= 43.34 ml =43.34/1000 liters

Partial pressure of oxygen = 781.35 torr

n= PV/RT = 781.35* (43.34/1000)/(294* 0.08206)=1.4 moles

from the reaction

2H2O2----> 2H2O+ O2

1 moles of oxygen requires two moles of Peroxide

1.4 moles give 1.4*2 =2.8 moles of peroxide

mass of peroxide = 2.8*(34.02)= 95.256 gms

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