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QUESTION

# In the following reaction, what mass of calcium carbonate, CaCO3, would be required to produce 881 L of carbon dioxide, CO2, measured at STP? (I dont understand where to begin...)

I got 3.93xx10^3 "kg CaCO"_3.

You didn't write a reaction. But you probably meant this:

"CaCO"_3(s) stackrel(Delta)(->) "CaO"(s) + "CO"_2(g)

where Delta indicates high heat, and "CaCO"_3 (calcium carbonate) has decomposed into "CaO"(s) (calcium oxide) and "CO"_2(g) (gaseous carbon dioxide).

You can start by assuming "CO"_2(g) is an ideal gas. At STP, we are at a temperature of 0^@ "C", or "273.15 K", and "1 atm" of pressure. We know the volume produced is "881 L" of gas.

(The volume of solid is negligibly small in comparison.)

We are not told anything else, so we have to determine the volume in bb"1 mol" (the molar volume) of ideal gas to figure out how many "mol"s of "CO"_2 we made.

Recall the following equation for ideal gases:

bb(PV = nRT) (ideal gas law)

where P, V, n, R, and T are

• pressure ("atm"),
• volume ("L"),
• bb"mol"s of ideal gas,
• the universal gas constant (we choose "0.082057 L"cdot"atm/mol"cdot"K" because we are using "1 atm" for pressure and "L" for volume),
• and temperature in "K",

respectively.

Rearrange to get the molar volume, V/n:

V/n = (RT)/P

= (("0.082057 L"cdot"atm/mol"cdot"K")("273.15 K"))/("1 atm")

~~ "22.414 L".

So, the "mol"s of "CO"_2 we have is:

881 cancel"L" xx "1 mol"/(22.414 cancel"L") ~~ "39.31 mols CO"_2

So, back-calculations will give the bb"mol"s of "CaCO"_3 needed, and therefore the mass using its molar mass.

39.31 cancel("mols CO"_2) xx ("1 mol CaCO"_3(s))/(cancel("1 mol CO"_2(g)))

~~ "39.31 mols CaCO"_3(g)

So, we have to have used this much mass:

color(blue)(m_("CaCO"_3(s))) = 39.31 cancel("mols CaCO"_3(s)) xx "100.088 g"/(cancel("1 mol CaCO"_3(s)))

~~ "3934.1 g"

~~ color(blue)(3.93xx10^3 "kg CaCO"_3)

where molar mass is just the sum of the atomic masses for each atom in the compound.