# Question 30d07

Mar 13, 2016

The volume of carbon dioxide is 1.31 L.

#### Explanation:

This is a double-barrelled question.

It is really a stoichiometry problem ("How much carbon dioxide?") combined with a Gas Laws problem ("What volume of carbon dioxide?").

The stoichiometry problem

What volume of carbon dioxide can be produced from 5.7 g of calcium carbonate?

The balanced equation is

${\text{CaCO"_3 + "2HCl" → "CaCl"_2 + "H"_2"O" + "CO}}_{2}$

${\text{Moles of CaCO"_3 = 5.7 color(red)(cancel(color(black)("g CaCO"_3))) × ("1 mol CaCO"_3)/(100.09 color(red)(cancel(color(black)("g CaCO"_3)))) = "0.0569 mol CaCO}}_{3}$

${\text{Moles of CO"_2 = 0.0569 color(red)(cancel(color(black)("mol CaCO"_3))) × ("1 mol CO"_2)/(1color(red)(cancel(color(black)( "mol CaCO"_3)))) = "0.0569 mol CO}}_{2}$

The Gas Law problem

What is the volume of "0.0569 mol CO"_2 at 30 °C and 1.08 atm?

This is a task for the Ideal Gas Law:

color(blue)(|bar(ul(color(white)(l)PV = nRTcolor(white)(l))|)

We can rearrange this to

color(blue)(|bar(ul(color(white)(l)V= (nRT)/Pcolor(white)(l))|)

$n = \text{0.0569 mol}$
$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$
$\text{T = (30 + 273.15) K = 303.15 K}$
$\text{P = 1.08 atm}$

Hence,

V = (0.0569 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"color(red)(cancel(color(black)("·atm·K"^"-1""mol"^"-1"))) × 303.15 color(red)(cancel(color(black)("K"))))/(1.08 color(red)(cancel(color(black)("atm")))) = "1.31 L"#

The volume of the ${\text{CO}}_{2}$ is 1.31 L.