Question

8) A candy bar contains 30.2 g of sucrose (cane sugar), C12H22011- When the candy bar is eaten, the sucrose is metabolized according to the overall equation: (unbalanced) C2H2201160 + 0219) → CO2 + H,00 Balance the equation, and find the mass of O, consumed and the masses of CO2 and H20 produced in this reaction.

Answer 1

1)

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

Balance C:

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

Balance H:

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

Balance O:

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

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

2)

Molar mass of C12H22O11,

MM = 12*MM(C) + 22*MM(H) + 11*MM(O)

= 12*12.01 + 22*1.008 + 11*16.0

= 342.296 g/mol

mass of C12H22O11 = 30.2 g

mol of C12H22O11 = (mass)/(molar mass)

= 30.2/3.423*10^2

= 8.823*10^-2 mol

According to balanced equation

mol of O2 required = (12/1)* moles of C12H22O11

= (12/1)*8.823*10^-2

= 1.059 mol

Molar mass of O2 = 32 g/mol

mass of O2 = number of mol * molar mass

= 1.059*32

= 33.88 g

Answer: 33.9 g

3)

According to balanced equation

mol of CO2 formed = (12/1)* moles of C12H22O11

= (12/1)*8.823*10^-2

= 1.059 mol

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

mass of CO2 = number of mol * molar mass

= 1.059*44.01

= 46.59 g

Answer: 46.6 g

4)

According to balanced equation

mol of H2O formed = (11/1)* moles of C12H22O11

= (11/1)*8.823*10^-2

= 0.9705 mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

mass of H2O = number of mol * molar mass

= 0.9705*18.02

= 17.48 g

Answer: 17.5 g