Question

Question #a4579

Answer 1

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

Explanation:

The balanced is equation

`color(white)(mmmm)"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 `"CO"_2` should
be 4.4 g.

We can use the Ideal Gas Law to calculate the mass of `"CO"_2`

`color(blue)(bar(ul(|color(white)(a/a)pV = nRTcolor(white)(a/a)|)))" "`

Since `n = "mass"/"molar mass" = m/M`, we can write the Ideal Gas Law as

`color(blue)(bar(ul(|color(white)(a/a)pV = m/MRTcolor(white)(a/a)|)))" "`

We can rearrange this to get

`m = (MpV)/(RT)`

NTP is defined as 1 atm and 20 °C.

In this problem,

`M = "44.01 g·mol"^"-1""`
`p = "1 atm"`
`V = "22 400 mL" = "22.4 L"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "(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.