# Question #cabe5

##### 2 Answers

I got

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 **temperature** of **pressure**. We know the **volume** produced is

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

We are not told anything else, so we have to determine the **volume in** *molar volume*) of ** ideal** gas to figure out how many

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 = (RT)/P#

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

#~~# #"22.414 L"# .

So, the

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

So, back-calculations will give the **s** of **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.

#### Explanation:

The balanced equation of reaction occurring for the production of

OR

In each case given above 1 mol or 22.4L

Taking atomic mass of

we get molar mass

So to produce 1 mol or 22.4L

Hence to produce 881L