Step 1. Start with the balanced chemical equation.
"2H"_2"O"_2 → "2H"_2"O" + "O"_22H2O2→2H2O+O2
(a) Mass of "H"_2"O"_2H2O2
"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".
