# Question cabe5

Oct 23, 2016

I got $3.93 \times {10}^{3} {\text{kg CaCO}}_{3}$.

You didn't write a reaction. But you probably meant this:

${\text{CaCO"_3(s) stackrel(Delta)(->) "CaO"(s) + "CO}}_{2} \left(g\right)$

where $\Delta$ indicates high heat, and ${\text{CaCO}}_{3}$ (calcium carbonate) has decomposed into $\text{CaO} \left(s\right)$ (calcium oxide) and ${\text{CO}}_{2} \left(g\right)$ (gaseous carbon dioxide).

You can start by assuming ${\text{CO}}_{2} \left(g\right)$ is an ideal gas. At STP, we are at a temperature of ${0}^{\circ} \text{C}$, or $\text{273.15 K}$, and $\text{1 atm}$ of pressure. We know the volume produced is $\text{881 L}$ 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 $\boldsymbol{\text{1 mol}}$ (the molar volume) of ideal gas to figure out how many $\text{mol}$s of ${\text{CO}}_{2}$ we made.

Recall the following equation for ideal gases:

$\boldsymbol{P V = n R T}$ (ideal gas law)

where $P$, $V$, $n$, $R$, and $T$ are

• pressure ($\text{atm}$),
• volume ($\text{L}$),
• $\boldsymbol{\text{mol}}$s of ideal gas,
• the universal gas constant (we choose $\text{0.082057 L"cdot"atm/mol"cdot"K}$ because we are using $\text{1 atm}$ for pressure and $\text{L}$ for volume),
• and temperature in $\text{K}$,

respectively.

Rearrange to get the molar volume, $\frac{V}{n}$:

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

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

$\approx$ $\text{22.414 L}$.

So, the $\text{mol}$s of ${\text{CO}}_{2}$ we have is:

$881 {\cancel{\text{L" xx "1 mol"/(22.414 cancel"L") ~~ "39.31 mols CO}}}_{2}$

So, back-calculations will give the $\boldsymbol{\text{mol}}$s of ${\text{CaCO}}_{3}$ 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)))

$\approx {\text{39.31 mols CaCO}}_{3} \left(g\right)$

So, we have to have used this much mass:

$\textcolor{b l u e}{{m}_{{\text{CaCO"_3(s))) = 39.31 cancel("mols CaCO"_3(s)) xx "100.088 g"/(cancel("1 mol CaCO}}_{3} \left(s\right)}}$

$\approx$ $\text{3934.1 g}$

$\approx$ $\textcolor{b l u e}{3.93 \times {10}^{3} {\text{kg CaCO}}_{3}}$

where molar mass is just the sum of the atomic masses for each atom in the compound.

Oct 23, 2016

$3.933 k g \text{ } C a C {O}_{3} \left(s\right)$

#### Explanation:

The balanced equation of reaction occurring for the production of $C {O}_{2} \left(g\right)$ may be any one of the following

$C a C {O}_{3} \left(s\right) + 2 H C l \left(a q\right) \to C a C {l}_{2} \left(a q\right) + C {O}_{2} \left(g\right) + {H}_{2} O$

OR

$C a C {O}_{3} \left(s\right) \stackrel{H e a t}{\to} C a O + C {O}_{2} \left(g\right)$

In each case given above 1 mol or 22.4L $C {O}_{2} \left(g\right)$ at STP is produced from 1 mole of $C a C {O}_{3} \left(s\right)$

Taking atomic mass of

$C a \to 40 \text{ g/mol}$

$C \to 12 \text{ g/mol}$

$O \to 16 \text{ g/mol}$

we get molar mass

$C a C {O}_{3} \to \left(40 + 12 + 3 \cdot 16\right) = 100 \text{ g/mol}$

So to produce 1 mol or 22.4L $C {O}_{2} \left(g\right)$ at STP we require 1 mol or 100g $C a C {O}_{3} \left(s\right)$

Hence to produce 881L $C {O}_{2} \left(g\right)$ at STP we require

(100g/"mol")/(22.4L/"mol")xx881L=3933g =3.933kg" "CaCO_3(s)#