Question

Question #47cfa

Answer 1

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)

Answer 2

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.