The volume of gas produced is 1.66 L.

There are four steps involved in this stoichiometry problem:

Step 1. Write the balanced chemical equation.

#M_r:color(white)(mll) 227.09#
#color(white)(mm)"4C"_3"H"_5"N"_3"O"_9"(l)" → underbrace("12CO"_2"(g)" + "10H"_2"O(g)" + "6N"_2"(g)" + "O"_2"(g)")_color(red)("29 mol of gas") #

Step 2. Convert grams of nitroglycerin to moles of nitroglycerin

#"moles of nitroglycerin" = 2.00 color(red)(cancel(color(black)("g nitroglycerin"))) × ("1 mol nitroglycerin")/(227.09 color(red)(cancel(color(black)("g nitroglycerin")))) = "0.008 807 mol nitroglycerin"#

Step 3. Convert moles of nitroglycerin to moles of gas

The molar ratio of gases to nitroglycerin is #"29 mol gas":"4 mol nitroglycerin"# .

∴ #"0.008 807"color(red)(cancel(color(black)("mol nitroglycerin"))) × ("29 mol gas")/(4 color(red)(cancel(color(black)("mol nitroglycerin")))) = "0.063 85 mol gas"#

4. Use the Ideal Gas Law to calculate the volume of gas.

The Ideal Gas Law is

#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#

where

#P# = the pressure of the gas,
#V# = the volume of the gas,
#n# = the number of moles of the gas,
#R# = the universal gas constant
#T# = the temperature of the gas

We can rearrange the Ideal Gas Law to get

#color(blue)(bar(ul(|color(white)(a/a)V = (nRT)/Pcolor(white)(a/a)|)))" "#

In this problem,

#n = "0.063 85 mol"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(33 + 273.15) K" = "306.15 K"#
#P = "0.968 atm"#

∴ #V = ("0.063 85" color(red)(cancel(color(black)("mol"))) × "0.082 06 L"· color(red)(cancel(color(black)("atm"·"K"^"-1""mol"^"-1"))) × 306.15 color(red)(cancel(color(black)("K"))))/(0.968 color(red)(cancel(color(black)("atm")))) = "1.66 L"#

The volume of gas is 1.66 L.

Here's a useful video on mass-volume conversions.

VIDEO