Question

16. Consider the combustion of solid sucrose (C12H22O11), to form liquid water and a gaseous product.
a) Write & balance the equation, including phases.
b)If 2.001 g of sucrose reacts with 1.00 L of O2 at 753 torr and 25.8 °C, how many
grams of liquid water are theoretically formed?
c) Assign oxidation numbers to each element in the reaction.
d) Which substance undergoes oxidation? Reduction?
e) Identify the oxidizing and reducing agents.

Answer 1

16)

a) C12H22O11(s) + O2(g) --------------------- CO2(g)   + H2O(l)

The balanced equation is

C12H22O11(s) + 12 O2(g) ---------------------   12 CO2(g)   + 11 H2O(l)

b)

    C12H22O11(s) + 12 O2(g) ---------------------   12 CO2(g)   + 11 H2O(l)

      1 mole                12 mole

mass of sucrose = 2.001 g

molar mass of sucrose = 342.3 gram/mole

number of moles of sucrose = mass /molar mass = 2.001 / 342.3 = 0.00584 moles

volume of O2 = 1.00L

Pressure of O2 = 753 torr = 0.991 atm             ( 1torr = 0.00131579 atm)

Temperature = T = 25.8C = 25.8+273 = 298.8 K

R = 0.0821 L-atm/mol-K

PV=nRT

n = PV/RT = 0.991 x 1.00 / 0.0821 x 298.8 = 0.0404 moles

number of moles of O2 = 0.0404 moles

according to equation

1 mole of C12H22O11 = 12 moles of O2

0.00584 moles of C12H22O11 = ?

                                              = 12 x 0.00584 / 1 = 0.07008 moles of O2

we need 0.07008 moles of O2, but we have 0.0404 moles. So , O2 is first completed during the reaction

Hence, O2 is limiting reagent

according to equation

12 moles of O2 = 11 moles of H2O

0.0404 moles of O2 = ?

                             = 0.0404 x 11/12 = 0.0370 moles of H2O

number of moles of H2O formed = 0.0370 moles

molar mass of H2O = 18.01 gram/mole

mass of one moles of H2O = 18.01 grams

mass of 0.0370 moles of H2O = 0.0370 x 18.01 = 0.666 grams

Theoritical mass of H2O = 0.666 grams

c)

C12H22O11(s) + 12 O2(g) ---------------------   12 CO2(g)   + 11 H2O(l)

0    +1   -2                  0                                  +4 -2            +1   -2

In Sucrose ,           Oxidation number of C = 0

                             Oxidation number of H = +1

                             Oxidation number of O = -2

Oxidation number of Oxygen in O2 =0

In CO2,

Oxidation number of C = +4

Oxidation number of O = -2

In H2O

Oxidation number of H = +1

Oxidation number of O = -2

d)

Oxidation number of Carbon is increases from 0 to +4. SO , C undergoes oxidation

Oxidation number of Oxygen is decreases from 0 to -2. So, O2 undergoes reduction

e) The reduced substance can acts as Oxidising agent

Hence , O2 acts as Oxidising agent

The oxidised substance can acts as reducing agent

Hence, Sucrose acts as reducing agent.