Question

Extra Credit Limiting Reactants and Ideal Gas Law 10 points Due Monday, 11/25/19, at the beginning of class In the limiting reactants lab you were asked to determine which reactant was limiting for each test tube and also to rank the size of each balloon filled with CO2 produced by the reaction: Na2CO3(s) + 2HC,H,O2 (aq) + CO2(g) + H2O(l) + 2NaCzH302 (aq) You have already calculated the number of moles of limiting reactants and moles of carbon dioxide produced from the limiting reactant from each test tube, fill that in below. THEN, using the ideal gas law, calculate the VOLUME of CO2 in each balloon. Use 0.9924 atm for the pressure and 21.6°C for the temperature. Finally, rerank the balloon size according to the volume you calculate. SHOW THE SETUP FOR YOUR VOLUME CALCULATIONS AND SIGNIFICANT FIGURES TO RECEIVE FULL first two require no math Test tube Formula of limiting reactant Moles of CO2 from limiting reactant Volume (L) of CO2 gas Rank 10.0027 ml 10.0055 HGH₂O 10.000 ml | HG, H₂O2 10.000ml | HGH₃02 0.008 * should already be done on your limiting reactants lab

Answer 1

Test tube #	Formula of limiting reactant*	Moles of CO2 from limiting reactant*	Volume (L) of CO2 gas	Rank
1	HC2H3O2	0.0027	15.18	Rank 1
2	HC2H3O2	0.0055	7.45	Rank 2
3	HC2H3O2	0.008	5.12	Rank 3
4	HC2H3O2	0.008	5.12	Rank 3
5	HC2H3O2	0.008	5.12	Rank 3

To calculate the volume we have use the Ideal gas equation , PV=nRT where P is pressure(given as 0.9924 atm), V is the volume (to be determined), n is the number of moles (given in the table), R is the gas constant (0.0821 atm L/mol K) and T is the temperature ( given as 21.6 degree Celcius = 294.75 K)

Thus, the calculation for each test tube are as follows

Test tube #1 : 0.9924 V₁= (0.0027)(0.0821)( 294.75)

Thus, V₁=[0.9924 / (0.0027)(0.0821)( 294.75)] = 15.18 L

Test tube #2: 0.9924 V₂= (0.0055)(0.0821)( 294.75)

Thus, V₂=[0.9924 / (0.0055)(0.0821)( 294.75)] = 7.45 L

For test tubes 3,4 and 5 , the calculations are the same since the

Test tube #3: 0.9924 V_3,4,5= (0.008)(0.0821)( 294.75)

Thus, V_3,4,5=[0.9924 / (0.008)(0.0821)( 294.75)] = 5.12L