Asked by Heidi
A butane lighter had a mass of 24.643 grams before lighting and a mass of 24.592 grams after. How many mL of CO2 were produced? Assume STP.
Answers
Answered by
DrBob222
2C4H10 + 13O2 ==> 8CO2 + 10H2O
24.643 initial mass
-24.592 after using
--------
00.051g mass butane used.
mols butane = grams/molar mass = estimated 0.00088 but you need to get a more accurate figures.
Using the coefficients in the balanced equation, convert mols butane to mols CO2. That's estimated 0.00088 x (8 mols CO2/2 mols C4H10) = ?
Then use PV = nRT and solve for L; convert to mL.
24.643 initial mass
-24.592 after using
--------
00.051g mass butane used.
mols butane = grams/molar mass = estimated 0.00088 but you need to get a more accurate figures.
Using the coefficients in the balanced equation, convert mols butane to mols CO2. That's estimated 0.00088 x (8 mols CO2/2 mols C4H10) = ?
Then use PV = nRT and solve for L; convert to mL.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.