Step 1. Start with the balanced chemical equation.
#"2H"_2"O"_2 → "2H"_2"O" + "O"_2#
(a) Mass of #"H"_2"O"_2#
#"Mass of H"_2"O"_2 = 1270 color(red)(cancel(color(black)("mL H"_2"O"_2))) × ("1.00 g H"_2"O"_2)/(1 color(red)(cancel(color(black)("mL H"_2"O"_2)))) = "1270 g H"_2"O"_2#
(b) Moles of #"H"_2"O"_2#
#1270 color(red)(cancel(color(black)("g H"_2"O"_2))) × ("1 mol H"_2"O"_2)/(34.01 color(red)(cancel(color(black)("g H"_2"O"_2)))) = "37.34 mol H"_2"O"_2 #
(b) Moles of #"O"_2#
#37.34 color(red)(cancel(color(black)("mol H"_2"O"_2))) × ("1 mol O"_2)/(2 color(red)(cancel(color(black)("mol H"_2"O"_2)))) = "18.67 mol O"_2#
(c) Volume of #"O"_2#
The Ideal Gas Law is
#color(blue)(bar(ul(|color(white)(a/a)PV = nRTcolor(white)(a/a)|)))" "#
We can rearrange this to give
#V = (nRT)/P#
#n = "18.67 mol O"_2#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(78 + 273.15) K" = "351.15 K"#
#P = "0.934 atm"#
∴ #V = (18.67 color(red)(cancel(color(black)("mol"))) × "0.082 06 L"·color(red)(cancel(color(black)("atm·K"^"-1""mol"^"-1")))
× 351.15 color(red)(cancel(color(black)( "K"))))/(0.934 color(red)(cancel(color(black)("atm")))) = "576 L"#
The volume of #"O"_2# produced is #"576 L"#.
