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# 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.