# Question 47cfa

May 1, 2015

The mass of ${\text{CO}}_{2}$ is 0.5497 g, and its volume at STP is 0.2800 L.

$\text{CaCO"_3("s")+"heat}$$\rightarrow$$\text{CaO(s)+CO"_2("g")}$

Determine the mole:mole ratio for ${\text{CaCO}}_{3}$ and ${\text{CO}}_{2}$.

$\left(1 \text{mol CaCO"_3)/(1 "mol CO"_2}\right)$ and $\left(1 {\text{mol CO"_2)/(1 "mol CaCO}}_{2}\right)$

Determine the molar mass of ${\text{CaCO}}_{3}$ and ${\text{CO}}_{2}$.

Molar mass of $\text{CaCO"_3=100.0869 "g/mol}$
Molar mass of $\text{CO"_2=44.009 "g/mol}$

Determine the number of moles of ${\text{CaCO}}_{3}$ in $1.25 \text{g CO"_2}$.

$1.25 \cancel{{\text{g CaCO"_3)xx(1 "mol CaCO"_3)/(100.0869 cancel("g CaCO}}_{3}}$ = $0.01249 \text{mol CaCO"_3}$

Determine the number of moles of ${\text{CO}}_{2}$.

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 ${\text{CO}}_{2}$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 ${\text{CO}}_{2}$ at STP.

Use the ideal gas law equation $P V = n R T$

STP for gases is $273.15 \text{Kelvins}$ and $1 \text{atm}$.

Given/Known:
$P = 1 \text{atm}$
$T = 273.15 \text{K}$
$n = 0.01249 \text{mol}$
$R = 0.08206 {\text{L atm K"^(-1) "mol}}^{- 1}$

Unknown:
$V$

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

$V = \frac{n R T}{P}$ =

$\left(0.01249 \text{mol" * 0.08206 "L atm K"^(-1) "mol"^(-1) * 273.15 "K")/(1 "atm}\right)$ =

$0.2800 \text{L CO"_2}$ (rounded to 4 significant figures)

May 4, 2015

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.