How many grams of oxygen gas is essentially required for complete combustion of 3 moles of butane gas ??
Answers
Answered by
13
2C4H10 + 13O2 ---> 8CO2 + 10H2O
Since, 13 moles of O2 is required for complete combustion of 2 moles of butane
Therefore, 1 mole butane require 13/2moles of
Therefore, 3 moles require 13/2×3 = 39/2 = 19.5 moles
We know that,
Number of moles = W(mass in grams)/M ( Molecular mass)
Thus, weight of O2 required = 19.5×32
= 624g
Since, 13 moles of O2 is required for complete combustion of 2 moles of butane
Therefore, 1 mole butane require 13/2moles of
Therefore, 3 moles require 13/2×3 = 39/2 = 19.5 moles
We know that,
Number of moles = W(mass in grams)/M ( Molecular mass)
Thus, weight of O2 required = 19.5×32
= 624g
Answered by
1
Answer:
I am not mood to the answer for me to get to the attachment file and the result is a good time to do in order for the brainliest the brainliest is your answer I have a b 2 1 2 1 the question is how the answer for you and your answer is no way of how you would be a great weekend the brainliest the answer
Similar questions