Question

11. Use the equation below to calculate the following. a. Balance the reaction. K2CO3 (aq) + _HCH3O2 (aq) → KCH:O2 (aq) + H20 (1) +_ CO2 (g) b. Moles of H2O that can be obtained from 25.0 mL of 0.150 M HC2H302 c. Volume of 0.210 MK2CO3 needed to produce 17.5 mol KC,H,O, d. Volume of 1.25 M HC,H,O2 needed to react with 75.2 mL 0.750 M K.CO; e. Liters of CO2 gas at 25°C and 0.750 atm, produced by the reaction of 105 mL of 1.50 M HC,H,O, f. Liters of CO2 gas at 35°C and 0.975 atm, produced by the reaction of 25.0 mL of 0.350 M K COand 25.0 mL of 0.250 M HC2H:O2 (Hint: you need to determine the limiting reactant)

Answer 1

111) a) K, Coz.caq yt 2. HC,11307.caq 3 —> 2 KC, H49, caq, + HOCA) +0,9) moly of HGH30, = 25.0 x0.150 - 3.75*10mol 1000 moles of H2O formed = 3.75x103 2 -3 1 31.875x10 mo moles of k&2 H3O2 = 17.5 moles of Kzcom - 17.5 - 8.75 0.210x V -8.75 V = 41.67 Volume of K, co, -41-67L moles of HC₂H302 moles of k2co₂ = 75.24 0.750. -0.0564 1000 V H2 H302 = 0.0564x2 = 0.1128 1.25x v=0.1128 v=0.090L Volume of HC₂H3 = 90.24m2 moles of HC₂ H₂O = 0.1575 moles of co₂ = 0.07875 mol pr=nRT 0.750 *V = 0.07875X0.08 21 x 298 vs 2.57L

f)

moles of K2CO3 = 25 x 0.350 / 1000 = 8.75 x 10^-3 mol

moles of HC2H3O2 = 25.0 x 0.250 / 1000 = 6.25 x 10^-3 mol

k2CO3   +      2 HC2H3O2   -----------> 2 KC2H3O2 + CO2   +   H2O

    1                   2                                                          1

8.75 x 10^-3     6.25 x 10^-3

here limiting reagent is HC2H3O2

moles of CO2 formed = 6.25 x 10^-3 / 2 = 3.125 x 10^-3

P V = n R T

0.975 x V = 0.003125 x 0.0821 x 308

V = 0.0810 L

volume of CO2 = 0.0810 L