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.
