Question #cabe5

2 Answers
Oct 23, 2016

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.

Oct 23, 2016

Answer:

#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)#