Answer:
V = 0.798 L
Explanation:
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In this case, for this gas stoichiometry problem, we first need to compute the moles of carbon dioxide via stoichiometry and the molar mass of starting calcium carbonate:
Next, we use the ideal gas equation for computing the volume, by bearing to mind that the STP conditions stand for a pressure of 1 atm and a temperature of 273.15 K:
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Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = = 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>