#"56.9 g WO"_3"# reacts with excess #"H"_2"# to produce #"W"# and #"H"_2"O"#. What is the percent yield of water if the actual yield is #"10.0 g H"_2"O"#?

Question

#"WO"_3 + "H"_2"##rarr##"W+3H"_2"O"#

Olivia Anton · Accepted Answer

The percent yield of water is #75%#.

Create a balanced equation first. #"WO"_3 + "3H"_2"##rarr##"W + 3H"_2"O"# The following steps must be followed in order to determine the expected (theoretical) mass of water in this reaction. The theoretical yield must be ascertained. This is accomplished by utilizing stoichiometry to determine the mass of water expected. #color(red)("given mass WO"_3"##rarr##color(blue)("mol WO"_3"##rarr##color(purple)("mol H"_2"O"##rarr##color(green)("mass H"_2"O"# It will be necessary to ascertain the molar masses of water and tungsten oxide. Molar Masses: Add the results for each substance to get the total. Multiply the subscript of each element by its molar mass, which is its atomic weight on the periodic table in g/mol. #"WO"_3:##(1xx183.84color(white)(.)"g/mol W")+(3xx15.999color(white)(.)"g/mol O")="231.837 g/mol WO"_3"# #"H"_2"O":##(2xx1.008color(white)(.)"g/mol H")+(1xx15.999"g/mol O")="18.015 g/mol H"_2"O"# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Mole Ratios Determine the mol ratio between #"WO"_3"# and #"H"_2"O"# from the balanced equation. #(1"mol WO"_3)/(3"mol H"_2"O")# and #(3"mol H"_2"O")/(1"mol WO"_3)# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(blue)("Moles Tungsten Oxide"# Multiply the #color(red)("given mass WO"_3"# by the inverse of its molar mass. #color(red)(56.9cancel"g WO"_3)xx(1"mol WO"_3)/(231.837color(red)cancel(color(black)("g WO"_3)))=color(blue)("0.2454 mol WO"_3"# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(purple)("Moles Water"# Multiply mol #"WO"_3# by the mole ratio with water in the numerator. #0.2454color(red)(cancel(color(black)("mol WO"_3)))xxcolor(black)(3"mol H"_2"O")/(color(red)(cancel(color(black)(1"mol WO"_3))))=color(purple)("0.7362 mol H"_2"O"# ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ #color(green)("Mass of Water"# Multiply mol water by its molar mass. #0.7362color(red)(cancel(color(black)("mol H"_2"O")))xx(18.015"g H"_2"O")/(color(red)(cancel(color(black)(1"mol H"_2"O"))))=color(green)("13.3 g H"_2"O")# rounded to three significant figures ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Theoretical Yield of Water is #"13.3 g"#. Actual yield is #"10.0 g H"_2"O"#. #"Percent Yield"=("actual yield")/("theoretical yield")xx100# #"Percent Yield"=("10.0 g H"_2"O")/("13.3 g H"_2"O")xx100="75.2%"# rounded to three significant figures

Avery Adams · Answer

undefined The molar mass of WO₃ is 231.84 g/mol and the molar mass of H₂O is 18.02 g/mol. From the balanced chemical equation, 1 mole of WO₃ produces 3 moles of H₂O. Therefore, the theoretical yield of H₂O can be calculated using stoichiometry.

1. Calculate the moles of WO₃:
$56.9 g \times \frac{1 mol}{231.84 g} = 0.2456 mol$

2. Calculate the moles of H₂O (using the stoichiometric ratio):
$0.2456 mol \times \frac{3 mol H₂O}{1 mol WO₃} = 0.7368 mol$

3. Calculate the theoretical yield of H₂O:
$0.7368 mol \times 18.02 g/mol = 13.27 g$

4. Calculate the percent yield of water:
$\text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$
$\text{Percent yield} = \frac{10.0 g}{13.27 g} \times 100 = 75.29\%$

Therefore, the percent yield of water is 75.29%.