Given the following, how many grams of water could be produced in this reaction? What is the limiting reactant? If 8.7 g of water are produced, what is the percent yield?

13.74 L of hydrogen gas at 30.0 C and 801 torr and 6.53 L of oxygen gas at 25 C and 801 torr are drawn into a cylinder where the following reaction takes place:

#2H_2(g) + O_2(g) -> 2H_2O(L)#

1 Answer
Ernest Z.
Sep 15, 2016

WARNING! Long Answer! The theoretical yield of water is 10.1 g.
#"O"_2# is the limiting reactant. The percent yield is 86 %.

Explanation:

We have to

  • use the Ideal Gas Law to calculate the moles of each gas
  • identify the limiting reactant
  • calculate the percent yield
  • calculate the theoretical yield

Moles of #"H"_2#

From the Ideal Gas Law,

#n = (PV)/(RT)#

#P = 801 color(red)(cancel(color(black)("torr"))) × "1 atm"/(760 color(red)(cancel(color(black)("torr")))) = "1.054 atm"#

#V = "13.74 L"#

#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#

#T = "(30.0 + 273.15) K" = "303.15 K"#

∴ #n = (1.054 color(red)(cancel(color(black)("atm"))) × 13.74 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 303.15 color(red)(cancel(color(black)("K")))) = "0.5821 mol"#

Moles of #"O"_2#

#P = "801 torr" = "1.054 atm"#
#V = "6.53 L"#
#R = "0.082 06 L·atm·K"^"-1""mol"^"-1"#
#T = "(25 + 273.15) K" = "298.15 K"#

∴ #n = (1.054 color(red)(cancel(color(black)("atm"))) × 6.53 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 298.15 color(red)(cancel(color(black)("K")))) = "0.2813 mol"#

Identify the limiting reactant

a. Calculate the moles of #"H"_2"O"# formed from the #"H"_2#

#"Moles of H"_2"O" = 0.5821 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.5821 mol H"_2"O"#

b. Calculate the moles of #"H"_2"O"# formed from the #"O"_2#

#"Moles of H"_2"O" = 0.2813 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.5626 mol H"_2"O"#

The limiting reactant is #"O"_2#, because it produces the fewest moles of product.

Calculate the theoretical yield of #"H"_2"O"#

#"Theoretical yield" = 0.5626 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")))) = "10.1 g H"_2"O"#

Calculate the percentage yield

#"% yield" = "actual yield"/"theoretical yield" × 100 % = (8.7 color(red)(cancel(color(black)("g"))))/(10.1 color(red)(cancel(color(black)("g")))) × 100 % = 86 %#

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