Question

1. A large quantity of chemical energy is stored in a gummy bear, which is primarily sugar, C12H22O11. When a gummy bear is placed in molten KClO3, the sugar is oxidized according to the following thermochemical equation: C12H22O11(s) + 8 KClO3(l) ⇒ 8 KCl(s) + 12 CO2(g) + 11 H2O(g) ∆Hrxn = −5488 kJ
If a gummy bear has an average mass of 0.5250 g and is 91.40% sugar, then how many whole gummy bears does it take to bring a cup of water (250.0 mL) at an initial temperature of 20.0°C to the boiling point? The boiling point of water is 100.0°C, the density of water (20°C) is 0.9982 g/mL, and the specific heat capacity of water is 4.184 J/(g•°C).

2. When 50.0 mL of 1.00 M Pb(NO3)2 (d = 1.30 g/mL) and 50.0 mL of 1.50 M KI (d = 1.18 g/mL) are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture increases from 19.0°C to 23.3°C.
(a) Indicate if the precipitation reaction is endothermic or exothermic.
(b) Write both the complete and net ionic chemical equations, including phase symbols, for the reaction that occurs when the two solutions are mixed.
complete:
net ionic:
(c) Calculate the moles of the limiting reactant.
(d) Calculate ∆Hrxn in kilojoules for the precipitation reaction, assuming that the heat capacity of the calorimeter is 15 J/°C in this case and the specific heat capacity of the solution after mixing is the same as that of water.

Answer 1

mass of water =volume* density = 250*0.9982 =249.55 gm

heat taken by water= mass* specific heat* temperature difference = 249.55*4.184*(100-20)=83530 joules

1 mole of sugar produces 5488 kj. molar mass of sugar (C12h22O11)= 12*12+22+16*11 = 342

342 gm produces 5488 Kj= 5488*1000 joules

0.525*0.914 gm produces 0.525*0.914*5488*1000/342 =7700 joules

so number of bags required = 83530/7700=11 bags

2. The reaction is Pb(NO3)2 +2KI ----->PbI2 + 2KNO3

1 mole of Pb(NO3)2 requires 2 mole of KI

molar ratio of Pb(NO3)2 : Ki =1:2

moles of Pb(NO3)2 = 1*50/1000 =0.05 moles

moles of KI= 50*1.5/1000 =0.075, actual retio = 0.05: 0.075 = 1 : 1.5

since KI is less than the required ratio of 2, it is the limiting reactant.

since there is temperature rise for water, the reaction is exothermic.

ionic equation Pb+2+ 2NO3-1+ 2K+ 2I- ----->PbI2 + 2K+ 2NO3-1

net ionic

Pb+2 (aq) + 2I- (aq)---------> PbI2 (s)

heat of reaction =- heat taken by water + heat taken by calorimeter

= mass * specific heat* twemperature difference + heat capacity of calorimeter* temperature difference

mass of mixture= 50*1.3+50*1.18 =124 gm

heat of reaction = 124*4.18*(23.3-19) + 15*(23.3-19)=2293 joules