Answer:
V = 0.798 L
Explanation:
Hello there!
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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0.00011811023622 <span>i</span><span>nches</span>
Answer:
option X^2- as member of group 2A are non metal