15 2) In this laboratory experiment you are to generate sufficie nea lculations

Question

15 2) In this laboratory experiment you are to generate sufficie nea lculations to plastic zipper type bag. You first must carry out some caleulath to determine the quantity of reagents needed. The carbon dioxide will be a solution of acetic acid (CH,COOH gencrated by reacting solid sodium bicarbonate (NaHCOs) with ). This is actually a two-step process that proceeds as reacting solid follows: Step l: NaHCO(s)+CHs Step 2: H.cofaq) H2O() + d0dg) You measure the plastic bag di thickness of bag to be? mensions to be 7.15 in. by 7.97 in. Assuming the average when inflated will be 1.50 in. what do you calculate the volume of the Of course, this To get a better volume of the water to be 1274.7 mL. Now, you measure room temperature to volume is only approximate, as the width of the inflated bag is not uniform. value for the volume, you fill the bag with water and then measure the be 21 °C, and the barometer on gives a reading of 756 torr for the atmospheric pressure Using the ideal gas law, calculate the number of moles of CO, that will be required to fill the bag at this temperature and pressure. Now, inspect the reactions above that are generating CO, and calculate the molar quantity of NaHCO, and CH,COOH needed Quantity of NaHCO: Quantity of CH COOH What mass of NaHCO, will be needed? You have a solution of 6.03 M acetic acid. What volume of this solution will be needed? 192

Explanation / Answer

Ans. #I. Volume of uniform bag = l x b x h = 7.15 in x 7.97 in x 1.50 in = 85.47825 in3

                                                            = 1.401 L                               ; [1 in3 = 0.0163871 L]

#II. Given,

            Actual volume of bag = 1274.7 mL = 1.2747 L

            Temperature, T = 21.00C = 294.15 K

            Pressure, P = 756 torr = (756 / 760) atm = 0.9947 atm

Ideal gas equation: PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol-1K-1

            T = absolute temperature (in K) = (0C + 273.15) K

Putting the values in equation 1-

0.9947 atm x 1.2747 L = n x (0.0821 atm L mol-1K-1) x 294.15 K

            Or, n = 1.26794409 atm L / (24.149715 atm L mol-1)

            Hence, n = 0.0525 mol

Therefore, required moles of CO2 = 0.0525 mol

#III. Note that-

            1 mol NaHCO3 (step 1 reaction) generates 1 mol H2CO3

            1 mol CH3COOH (step 1) generates 1 mol H2CO3

            1 mol H2CO3 (step 2 reaction) generates 1 mol CO2

            That is, 1 mol NaHCO3 and 1 mol CH3COOH generates 1 mol CO2

So,

            Required moles of NaHCO3 = 0.0525 mol = moles of CO2 generates

Required moles of CH3COOH = 0.0525 mol = moles of CO2 generates

#IV. Required mass of NaHCO3 = Required moles x Molar mass

                                                = 0.0525 mol x (84.006908 g/ mol)

                                                = 4.410 g

#V. Required volume of acetic acid = Required moles / Molarity

                                                = 0.0525 mol / (6.03 M)                   ; [1 M = 1 mol/ L]

                                                = 0.0525 mol / (6.03 mol/ L)

                                                = 0.008706 L

                                                = 8.706 mL

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