#2H_2O - 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?

Question

#2H_2O -> 2H_2 + O_2#  What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?

Wyatt Atkins · Accepted Answer

#67.5%#

The chemical equation that is in balance tells you this.

#2"H"_ 2"O"_ ((l)) -> 2"H"_ (2(g)) + "O"_ (2(g))#

that it takes #2# moles of water to produce #1# mole of oxygen gas. Use the molar masses of the two chemical species to convert this mole ratio to a gram ratio

#("2 moles H"_2"O")/("1 mole O"_2) = (2 color(red)(cancel(color(black)("moles H"_2"O"))) * ("18.015 g H"_2"O")/(1color(red)(cancel(color(black)("mole H"_2"O")))))/(1color(red)(cancel(color(black)("mole O"_2))) * "32.0 g O"_ 2/(1color(red)(cancel(color(black)("mole O"_2))))) = ("36.03 g H"_2"O")/("32.0 g O"_2)#

So, you know that every #"36.03 g"# of water that take part in the reaction produce #"32.0 g"# of oxygen gas.

This means that when #"17.0 g"# of water undergo decomposition, you should expect to get

#17.0 color(red)(cancel(color(black)("g H"_2"O"))) * "32.0 g O"_2/(36.03color(red)(cancel(color(black)("g H"_2"O")))) = "15.1 g O"_2#

This represents the reaction's theoretical yield, i.e. what is produced by a reaction that has a #100%# yield.

Now, you know that the actual yield of the reaction is #"10.2 g"# of oxygen gas. To find the percent yield, divide the actual yield by the theoretical yield and multiply the result by #100%#.

You ought to obtain

#"% yield" = (10.2 color(red)(cancel(color(black)("g O"_2))))/(15.1color(red)(cancel(color(black)("g O"_2)))) * 100% = color(darkgreen)(ul(color(black)(67.5%)))#

Three sig figs are used to round the result.

Jeremiah Adams · Answer

undefined To find the percent yield of O₂, we first need to determine the theoretical yield of O₂ and then use it to calculate the percent yield.

1. Calculate the molar mass of H₂O:
   - H: 1.008 g/mol
   - O: 16.00 g/mol
   Molar mass of H₂O = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

2. Determine the number of moles of H₂O:
   Moles of H₂O = 17.0 g / 18.02 g/mol ≈ 0.943 mol

3. Use the balanced chemical equation to find the theoretical yield of O₂:
   From the balanced equation, 2 moles of H₂O produce 1 mole of O₂.
   Moles of O₂ = 0.943 mol × (1 mol O₂ / 2 mol H₂O) = 0.472 mol

4. Calculate the theoretical yield of O₂ in grams:
   The molar mass of O₂ is 32.00 g/mol.
   Theoretical yield of O₂ = 0.472 mol × 32.00 g/mol = 15.10 g

5. Calculate the percent yield of O₂:
   Percent yield = (actual yield / theoretical yield) × 100%
   Percent yield = (10.2 g / 15.10 g) × 100%
   Percent yield ≈ 67.55%

Therefore, the percent yield of O₂ is approximately 67.55%.

#2H_2O -&gt; 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?

#2H_2O -> 2H_2 + O_2# What is the percent yield of #O_2# if 10.2 g of #O_2# is produced from the decomposition of 17.0 g of #H_2O#?