Question

2.Q.4. Find weight of O2 needed to burn
14gm ethene
(C2H4) using 32 gms
of oxygen
​

Answer 1

Answer:

...48gm...

use the formula CxHy +(X+y/4)O2= xCO2 +y/4H20

Explanation:

So as question given

C2H4 + O2 = CO2+H2O

Now balance it according to above equation

C2H4 + (2+4/4)O2= 2CO2 + 4/2O2

we get equation is equal to=

C2H4+ 3O2 = 2CO2 + 2H20

Now the question says 14gm ethene means 14/28 0.5 mole

So if one mile C2H4 requires the 3 mole of O2 so 1/2 mole require = 3/2 mole

so weight required to burn = 32×3/2=48gm...