When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s) + CO₂(g) What is the mass of calcium carbonate needed to produce 35.0 L of carbon dioxide at STP? Express your answer with the appropriate units. View Available Hint(s) 0 mass of CaCO3 = 370.8 Part B μᾶ atm Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining 2 Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10 (9) + 1302 (g)→8CO2 (g) + 10H₂O(1) At 1.00 atm and 23 °C, what is the volume of carbon dioxide formed by the combustion of 2.20 g of butane?
When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction CaCO3(s)→CaO(s) + CO₂(g) What is the mass of calcium carbonate needed to produce 35.0 L of carbon dioxide at STP? Express your answer with the appropriate units. View Available Hint(s) 0 mass of CaCO3 = 370.8 Part B μᾶ atm Submit Previous Answers Request Answer Incorrect; Try Again; 4 attempts remaining 2 Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2C4H10 (9) + 1302 (g)→8CO2 (g) + 10H₂O(1) At 1.00 atm and 23 °C, what is the volume of carbon dioxide formed by the combustion of 2.20 g of butane?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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
Transcribed Image Text:When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction
CaCO3(s)→CaO(s) + CO₂(g)
What is the mass of calcium carbonate needed to produce 35.0 L of carbon dioxide at STP?
Express your answer with the appropriate units.
View Available Hint(s)
mass of CaCO3
Submit
=
Part B
0
370.8
HÅ
atm
Previous Answers Request Answer
X Incorrect; Try Again; 4 attempts remaining
?
Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete
combustion of butane is
2C4H10 (9) + 1302 (g)→8CO2 (g) + 10H₂O (1)
At 1.00 atm and 23 °C, what is the volume of carbon dioxide formed by the combustion of 2.20 g of butane?
Express your answer with the appropriate units.

Transcribed Image Text:The ideal gas law
PV = nRT
relates pressure P, volume V, temperature T, and
number of moles of a gas, n. The gas constant R
equals 0.08206 L.atm/(K · mol) or 8.3145
J/(K. mol). The equation can be rearranged as
.
follows to solve for n:
PV
RT
n =
This equation is useful when dealing with gaseous
reactions because stoichiometric calculations involve
mole ratios.
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