GATE Civil Engineering (CE) 2020 Shift 2 Solved Paper

‌

C

(12)

+

O(2)

(32)

⟶

CO2

(44)

Oxygen required for 1t/ day =‌32/12=2.6667 moles
Amount of air required for combustion of carbon =2.6667×350=933.33 tonnes/day Similarly,
for hydrogen   ‌

32

4

×100=800
for Sulphur   ‌

32

32

×60=60
for nitrogen   

32

14

×30=68.57
∴ total oxygen required Available amount of oxygen =260t/day
∴ required amount of oxygen =1862−260=1602t/day
∴ Amount of air required =1602/.23=6965.2 tonnes/day