Question

Question #cabe5

Answer 1

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

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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.

Answer 2

`3.933kg" "CaCO_3(s)`

Explanation:

The balanced equation of reaction occurring for the production of `CO_2(g)` may be any one of the following

`CaCO_3(s)+2HCl(aq)->CaCl_2(aq)+CO_2(g)+H_2O`

OR

`CaCO_3(s)stackrel(Heat)->CaO+CO_2(g)`

In each case given above 1 mol or 22.4L `CO_2(g)` at STP is produced from 1 mole of `CaCO_3(s)`

Taking atomic mass of

`Ca->40" g/mol"`

`C->12" g/mol"`

`O->16" g/mol"`

we get molar mass

`CaCO_3->(40+12+3*16)=100" g/mol"`

So to produce 1 mol or 22.4L `CO_2(g)` at STP we require 1 mol or 100g `CaCO_3(s)`

Hence to produce 881L `CO_2(g)` at STP we require

`(100g/"mol")/(22.4L/"mol")xx881L=3933g =3.933kg" "CaCO_3(s)`