Question #47cfa

The mass of #"CO"_2# is 0.5497 g, and its volume at STP is 0.2800 L.

Start with a balanced equation, which you have.

#"CaCO"_3("s")+"heat"##rarr##"CaO(s)+CO"_2("g")"#

Determine the mole:mole ratio for #"CaCO"_3# and #"CO"_2#.

#(1 "mol CaCO"_3)/(1 "mol CO"_2")# and #(1 "mol CO"_2)/(1 "mol CaCO"_2)#

Determine the molar mass of #"CaCO"_3# and #"CO"_2#.

Molar mass of #"CaCO"_3=100.0869 "g/mol"#
Molar mass of #"CO"_2=44.009 "g/mol"#

Determine the number of moles of #"CaCO"_3# in #1.25 "g CO"_2"#.

#1.25 cancel("g CaCO"_3)xx(1 "mol CaCO"_3)/(100.0869 cancel("g CaCO"_3)# = #0.01249 "mol CaCO"_3"#

Determine the number of moles of #"CO"_2#.

#0.01249 cancel("mol CaCO"_3)xx(1 "mol CO"_2)/(1 cancel("mol CaCO"_3))=0.01249 "mol CO"_2#

Determine the mass of #"CO"_2#by multiplying moles times molar mass.

#0.01249 cancel("mol CO"_2)xx(44.009 "g CO"_2)/(1 cancel("mol CO"_2))=0.5497 "g CO"_2#

Determine the volume of #"CO"_2# at STP.

Use the ideal gas law equation #PV=nRT#

STP for gases is #273.15 "Kelvins"# and #1 "atm"#.

Given/Known:
#P=1 "atm"#
#T=273.15 "K"#
#n=0.01249 "mol"#
#R=0.08206 "L atm K"^(-1) "mol"^(-1)#

Unknown:
#V#

Now rearrange the ideal gas equation in order to isolate and solve for #V#.

#V=(nRT)/(P)# =

#(0.01249 "mol" * 0.08206 "L atm K"^(-1) "mol"^(-1) * 273.15 "K")/(1 "atm")# =

#0.2800 "L CO"_2"# (rounded to 4 significant figures)

CaCO3(s) --> CaO(s) + CO2(g)

The first step to this problem is to find the number of moles that can be obtain from 1.25g.

Formula: (mass/relative molecular mass)

You will obtain an amount of 0.0125 moles. Your equation should be (1.25g/100)

STP (Standard Temperature Pressure) states the volume of a gas when it's at a standard temperature of 273K and 1atm. At STP, one mole of gas occupies 22.4 L of molar volume.

Therefore, multiply 0.0125 moles of CaCO3 with 22.4 L of molar volume at STP. Your final answer should then be 0.2800L of CO2.