Identify the limiting reactant
#"2H"_2 + "O"_2 → "2H"_2"O"#
#"Moles of H"_2 = 0.6 color(red)(cancel(color(black)("g H"_2))) × ("1 mol H"_2)/(2.02 color(red)(cancel(color(black)("g H"_2)))) = "0.297 mol H"_2#
#"Moles of H"_2"O from H"_2 = 0.297 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.297 mol H"_2"O"#
#"Moles of O"_2 = 11.2 color(red)(cancel(color(black)("L O"_2))) × ("1 mol O"_2)/(22.4 color(red)(cancel(color(black)("L O"_2)))) = "0.500 mol O"_2#
#"Moles of H"_2"O from O"_2 = 0.500 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol O"_2)))) = "1.00 mol H"_2"O"#
#"H"_2# forms the smaller amount of #"H"_2"O"#, so #"H"_2# is the limiting reactant.
A. Calculate the mass of water produced
#"Mass of water" = 0.297 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "5.35 g H"_2"O"#
B. Calculate the volume of unused reactant at STP
#"Moles of O"_2 " used" = 0.297 color(red)(cancel(color(black)("mol H"_2))) × (1 "mol O"_2)/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.149 mol O"_2#
#"Moles of unused O"_2 = "0.500 mol - 0.149 mol" = "0.351 mol"#
#"Volume of unused O"_2 " at STP" = 0.351 color(red)(cancel(color(black)("mol"))) × ("22.4 L")/(1 color(red)(cancel(color(black)("mol")))) = "7.86 L"#
Note: You are using STP as 1 atm and 0 °C.
In 1982, the IUPAC established STP as 1 bar and 0 °C.
Under the new definition of STP, the molar volume of a gas is 22.7 L.
