The volume of oxygen required to burn 12.00 L ethane is calculated as follows
find the moles of C2H6 used
At STP 1 mole is always = 22.4 L, what about 12.00 L
= ( 12.00L x 1 moles) 22.4 L = 0.536 moles
write the reacting equation
2C2H6+ 7O2 = 4CO2 + 6H2O
by use of mole ratio between C2H6 :O2 which is 2:7 the moles of O2
= 0.536 x7/2= 1.876 moles
again at STP 1mole = 22.4 L what about 1.876 moles
= 22.4 L x 1.876 moles/ 1 mole = 42.02 L
Answer:
It should be 115.5 amu but you didnt type that as an answer choice so...
Explanation:
Bruh all you had to do was add oxygen and sulfur and subtract from the total to get your answer :D.
Answer:
it gets rust and could not run properly
the energy required to raise one kilogram of the substance one degree
