Question

Revised;
A gummybearcan be placed in a test tube of hot potassium chlorate. This is a two step process. Assume 1gummy bears mass us 2.5 grams and is only sucrose.
step 1:
__KClO3 ->___KCl(s) + __O2(g)
step 2:
__C12H22O11(s)+ __O2(g) -> __ CO2(g)+__H2O(l)
deltaHr = -5645 kJ/mol-rxn

c. Calculate the energy evolved as heat in the reaction shown when 1 gummy is burned.
d. Write the thermochemical equation for the combustion of the gummy.

Answer 1

In Balanced chemical reaction, atoms on both side of reaction are same.

Balanced equation are:

Step !

2KClO₃2 KCl (s) + 3O₂(g)

Step 2

C₁₂H₂₂O₁₁(s) + 12O₂(g) 12 CO₂(g) + 11H₂O (l) H_r = -5645 kJ/mol

(c) Mass of gummy = 2.5 g

Molar mass of sucrose = 342 g/mol

Moles of sucrose in 1 gummy = mass/ molar mass = 2.5 g/342g/mol = 0.00731 mol

From step 2, 1 mol of sucrose produces 5645 kJ of heat

So, 0.00731 mol of sucrose produces 41.26 kJ of heat

Thus, H_r = -41.26 kJ

(d) Thus reaction is:

0.00731 C₁₂H₂₂O₁₁(s) + 0.0877O₂(g) 0.0877 CO₂(g) + 0.0804H₂O (l) H_r = -41.26 kJ/mol