Question #30d07

1 Answer
Mar 13, 2016

Answer:

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

#"CaCO"_3 + "2HCl" → "CaCl"_2 + "H"_2"O" + "CO"_2#

#"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#

#"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 = "0.0569 mol"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#"T = (30 + 273.15) K = 303.15 K"#
#"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 #"CO"_2# is 1.31 L.