Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 20.4 g of carbon were burned in the presence o...Asked by Eric
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 26.4 g of carbon were burned in the presence of 82.1 g of oxygen, 11.7 g of oxygen remained unreacted. What mass of carbon dioxide was produced?
Express your answer to one decimal place and include the appropriate units.
Express your answer to one decimal place and include the appropriate units.
Answers
Answered by
Stephen Curry
When carbon is burned in air, it reacts with oxygen to form carbon dioxide. When 27.6 g of carbon...?
Answered by
Stephen Curry
C(s) + O2(g) --> CO2(g)
27.6g .... 88.0g ................. initial amounts
0g .........14.4g ................. amounts when reaction complete
That means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 27.6 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.
27.6g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) = 101.2 g CO2 or 101 grams of carbon to three significant digits.
27.6g .... 88.0g ................. initial amounts
0g .........14.4g ................. amounts when reaction complete
That means that C was the limiting reactant, and the amount of CO2 is based on the amount of carbon that burned. Covert 27.6 grams of carbon to moles. The moles of CO2 will be the same, since they are in a 1:1 mole ratio. Then convert the moles of CO2 to grams.
27.6g C x (1 mol C / 12.0 g C) x (1 mol CO2 / 1 mol C) x (44.0g CO2 / 1 mol CO2) = 101.2 g CO2 or 101 grams of carbon to three significant digits.
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.