One brand anti-acid chewing gum contains, Mass of chewing gum, Construct an equation for the complete oxidation, A typical motor car requires
PROBLEM:
One brand anti-acid chewing gum contains 2.5%
by mass of urea, (NH2)2CO. Each mole of urea reacts with
two moles of aqueous acid in the mouth to give CO2 and an ammonium
salt. Construct a balanced chemical equation for the reaction between urea and
the ethanoic acid, CH3COOH. By using the chemical equation calculate
the mass of chewing gum required to neutralize one gram of ethanoic acid. Given
that,
Mr of urea, (NH2)2CO=60
Mr of ethanoic acid, CH3COOH
= 60
SOLUTION:
Number of moles of ethanoic acid in
one gram = 1 / 60
= 0.0166 moles.
Mole ratio between;
(NH2)2CO : CH3COOH
1 : 2
Therefore, number of moles of urea =
0.0166 / 2
= 8.33 x 10-3 moles
Mass
of urea = 8.33 x 10-3 x 60
= 0.5 grams.
Mass of
chewing gum;
2.5g urea is present in
= 100g chewing gum.
0.5g urea is present in
= 100 / 2.5 x 0.5 chewing gum.
=
20g
PROBLEM:
By using the data
in this equation, estimate how much oxygen a person requires on average in a
life time and how far a moter car can go using the same amount of oxygen. The
typical daily food requirement of a person can be considered to be 1.2 Kg of
carbohydrate. The person obtains energy by the oxidation of the carbohydrate,
which can be represented by the formula (CH2O)n.
(a) (i)
Construct an equation for the
complete oxidation (combustion) of the
carbohydrate (CH2O)n.
(ii) The
empirical relative formula mass of the carbohydrate is 30. Use your equation
above to calculate the number of moles of oxygen required by the person each
day.
(iii) How many moles of oxygen will a person
require in a life-time of 70 years?
(a)
A typical motor car requires 6 dm3 of octane, C8H18 to travel 100Km.
(i)
Construct an equation for the complete combustion of octane.
(ii)
The density of octane is 0.7g cm-3. Calculate how many moles of
octane are present in the 6 dm3 of octane. [ Mr of octane, C8H18
= 114 ]
(iii)
Calculate how many moles of oxygen would be required to burn the 6
dm3 octane completely.
(b)
Calculate how many kilometers the car can travel using the same
amount of oxygen a person uses in a
life-time (your answer to (a) (iii) )
(Nov. 93 / P3 / Q.1)
SOLUTION:
(a) (i)
(ii) n = Mr / empirical formula mass
n = Mr / 30
Mr = 30n
Number of moles of carbohydrate =
1200 / 30n
= 40 / n moles.
Mole ratio between; (CH2O)n :
O2
1 : n
Therefore;
Number of moles of oxygen required
per day = 40 / n x n
= 40 moles.
(iv)
Number of moles of oxygen required
In a life-time of 70 year = 40 x (70 x 365)
=
1.022 x 106 moles
(b) (i)
(l) (g)
(g) (l)
(ii) Density = mass / volume
Mass =
density x volume (6dm3
= 6000 cm3)
= 0.7 x 6000
= 4200 grams of octane
Mole = mass
/ Mr
= 4200 / 114
= 36.84 moles of octane.
(iii) Mole ratio between;
Octane
: Oxygen
1
: 25/2
Therefore,
number of moles of oxygen
Required to
burn 6dm3 of octane = 36.84 x 25/2
= 460.5 moles.
(c) To
run 100 Km motor car, 460.5 moles of oxygen required.
By using 1.022 x 106
moles of oxygen;
460.5 moles of oxygen =
100 Km
1 mole of
oxygen = 100/460.5
1.022 x 106
moles of oxygen = 100/460.5 x 1.022 x 106
= 2.22 x 105 Km
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