Engineering Chemistry S Datta

406 ENGINEERING CHEMISTRY
Example 4. An ultimate analysis of 1 g coal for nitrogen (N) estimation is the Kjelldahl
N
method, the evolved NH 3
I
has collected in 25 ml
HG
10 K J H 2
SO 4
acid solution. To neutralise the
excess acid, 15 ml of 0.1 (N) NaOH was required. Calculate the % of nitrogen in the given
sample.
Sol. In the Kjelldahl method organic compounds containing nitrogen are heated with
concentrated H 2
SO 4
to convert to (NH 4
) 2
SO 4
. This (NH 4
) 2
SO 4
during soiling with alkali liberates
F N
NH 3
that is absorbed in H 2
SO 4 10
F
HG I K J . Excess acid is titrated with NaOH.
The volume of H 2
SO 4
consumed by NH 3
(25 ×.1 – 15 × .1) ml (N) = 1 ml (N)
Now, 100 c.c(N) H 2
SO 4
≡ 17 g NH 3
≡ 14 g N.
14
∴ 1 c.c(N) H 2
SO 4
= gN = 0.014 gN
1000
Again 0.014 g Nitrogen is present in 1 g coal.
So, % of N in the coal sample = 0.014 × 100 = 1.4%.
1
Example 5. A gaseous fuel has the following composition of volume:
H 2
= 24%; CH 4
= 30%; C 2
H 6
= 11%,
C 2
H 4
= 4.5%; C 4
H 8
= 2.5%; CO = 6%; CO 2
= 8%;
O 2
= 2% and N 2
= 12%.
Calculate (i) air to fuel ratio and (ii) volume of dry products of combustion using 40%
excess air.
Sol. Basis: 1 m 3 of gaseous fuel.
Volume of
Volume of products
combustible Reaction Volume of O 2
m 3 of combustion
gases (m 3 ) on dry basis (m 3 )
H 2
= 0.24 H 2
+ 1 2 O 2 = H 2
O 0.24 × 1 2 = 0.12 –
CH 4
= 0.30 CH 4
+ 2O 2
= CO 2
+ 2H 2
O 0.3 × 2 = 0.60 CO 2
= 0.3
C 2
H 6
= 0.11 C 2
H 6
+ 3 1 2 O 2 = 2CO 2
+ 3H 2
O 0. 1 × 7 2 = 0.385 CO 2
= 0.11 × 2 = 0.22
C 2
H 4
= 0.045 C 2
H 4
+ 3O 2
= 2CO 2
+ 2H 2
O 0.045 × 3 = 0.135 CO 2
= 0.045 × 2 = 0.09
C 4
H 8
= 0.025 C 4
H 8
+ 6O 2
= 4CO 2
+ 4H 2
O 0.025 × 6 = 0.15 CO 2
= 0.025 × 4 = 0.1
CO = 0.06 CO + 1 2 O 2 = CO 2
0.06 × 1 2 = 0.03 CO 2
= 0.06 × 1 = 0.06
Total = 1.42 m 3 Total = 0.77 m 3
Less = 0.02 m 3 CO 2
in fuel = 0.08 m 3
Net need Net CO 2
= 0.85 m 3
= 1.4 m 3 (O 2
)