Topic: Molar Volume of a Gas 1)Why was it necessary to subtract the volume of H2

Question

Topic: Molar Volume of a Gas 1)Why was it necessary to subtract the volume of H2O2 solution injected into the generator from the volume of the gas in the gas collection tube? Explain 2) Would the amount of gas collected in the gas collection tube have been higher,lower or the same in 4 g of MnO2 were added to the reaction tube instead of 2g? Explain 3) At the end of your experiment what substance are left in your reaction tube? 4) A student performs this experiment using 5.2% H2O2 solution, Detemine the molar volume of the collected gas. 3.0 ml of peroxide solution in syringe, 58.9 ml gas collected, room temp 23 degree C, barometric pressure in lab 760.6 mm Hg, Vapor pressure of water at 23 degree C = 21.07mm Hg

Explanation / Answer

Since there is no backgroung about the experiment, I am assuming this has to do with the decomposition of H2O2 in the presence of MnO2 catalyst into O2 and H2O

1) When H2O2 is injected a certain volume of air also gets displaced as a result the total volume would be the volume of air displaced + volume of gas produced. Subtracting the volume of H2O2 will give the net volume of gas produced.

2) Since MnO2 is a catalyst, the net amount of gas produced at equilibrium would not change relative to the amount of the catalyst used.

3) Reaction tube will have the catalyst, MnO2

4) P gas = Barometric pressure - vapor pressure of H2O = 760-21.07 = 738.93 mmHg = 0.972 atm

V gas at STP = {(Vmeasured gas * P gas)/T(room temp)} * 273 K/1 atm = 0.058 * 0.972* 273/(23+273) = 0.052L

given 5.2% H2O2 solution i.e 5.2 g of H2O2 /100ml

Vol of H2O2 taken = 3.0 ml, there amt of H2O2 = 3*5.2/100 = 0.156 g

# moles of H2O2 = wt H2O2/MOL WT = 0.156/34 = 0.0046 moles

As per the decompositions reaction:

2H2O2 ------------- 2H2O + O2

# O2 = #H2O2/2 = 0.0046/2 = 0.0023 MOLES

molar volume of gas = V (STP)/moles = 0.052/0.0023 = 22.6 L/mol

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