# Question a4579

May 6, 2017

Your result is not in accordance with the Law of Conservation of Mass.

#### Explanation:

The balanced is equation

$\textcolor{w h i t e}{m m m m} {\text{CaCO"_3 → "CaO" + "CO}}_{2}$
"Mass/g:" color(white)(ml)10color(white)(mmmll)5.6color(white)(mml)?

If the reaction obeys the Law of Conservation of Mass, the mass of ${\text{CO}}_{2}$ should
be 4.4 g.

We can use the Ideal Gas Law to calculate the mass of ${\text{CO}}_{2}$

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = n R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

Since $n = \text{mass"/"molar mass} = \frac{m}{M}$, we can write the Ideal Gas Law as

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} p V = \frac{m}{M} R T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this to get

$m = \frac{M p V}{R T}$

NTP is defined as 1 atm and 20 °C.

In this problem,

$M = \text{44.01 g·mol"^"-1}$
$p = \text{1 atm}$
$V = \text{22 400 mL" = "22.4 L}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$T = \text{(20 + 273.15) K" = "293.15 K}$

m = (44.01 "g"·color(red)(cancel(color(black)("mol"^"-1"))) × 1 color(red)(cancel(color(black)("atm"))) × 22.4 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1""mol"^"-1"))) × 293.15 color(red)(cancel(color(black)("K")))) = "41.0 g"#

This is quite different from the expected value of 4.4 g.

Your result does not support the Law of Conservation of Mass.