Question

b. When the volume of the flask is determined, the mass of the air is not important. Explain why. c. Explain why we consider the mass of the air in the"empty" flask when we are calculating the mass of the flask(Equation 8), but we do not consider the mass of air when calculating the volume of CO2(Equation 6). 2. At STP, what volume of CO2 could be generated from 3.0 g of marble chips? 3. At ordinary lab conditions of 20°C and 710 torr, what volume of co2 could be generated from 3.0 g of marble chips? From 30 g of marble chips?

Answer 1

2).   Mrable chips is CaCO3

                         CaCO3 --------- CaO + CO2

                      Mass of marble chip = 3.0 grams

                       Molar mass of marble chip = 100 grams/mole

                 Number of moles of CaCO3 = 3.0/100 = 0.03 moles

According to equation ,   1mole of CaCO3 = 1mole of CO2

                            for 0.03 mole of CaCO3 = 0.03 mole of CO2

so , the number of moles of CO2 = 0.03 moles

At STP, Volume of one of mole of CO2 = 22.4 L

            Volume of 0.03 mole of CO2 = 22.4x0.03 = 0.672 L

Volume of CO2 at STP = 0.672 L

3).   T,= 20C = 20 + 273 = 293K

      P = 710 torr = 0.9342 atm

    Mass of Marble chip = m = 3.0grmas

Molar mass of Marble chip = CaCO3 = M = 100 grams/mole

Gas constant = R= 0.0821 L-atm/mol/K

          Ideal gas equation =           PV=nRT

             but n= mass/molar mass

                      PV=nRT

                  PV= mRT/M

           V= mRT/PM

              V= 3.0x0.0821x293/0.9342x100 = 0.772 L

    Volume of CO2 = 0.772L.