# "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"?

## $\text{WO"_3 + "H"_2}$$\rightarrow$$\text{W+3H"_2"O}$

Apr 21, 2017

The percent yield of water is 75%.

#### Explanation:

$\text{WO"_3 + "3H"_2}$$\rightarrow$$\text{W + 3H"_2"O}$

The theoretical yield must be determined. This is done by determining the mass of water expected using stoichiometry. The following represents the steps that must be taken to determine the expected (theoretical) mass of water in this reaction.

color(red)("given mass WO"_3"$\rightarrow$color(blue)("mol WO"_3"$\rightarrow$color(purple)("mol H"_2"O"$\rightarrow$color(green)("mass H"_2"O"

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The molar masses of tungsten oxide and water will need to be determined.

Molar Masses
Multiply the subscript of each element by its molar mass, which is its atomic weight on the periodic table in g/mol. Add the results for each substance to get the total.

${\text{WO}}_{3} :$(1xx183.84color(white)(.)"g/mol W")+(3xx15.999color(white)(.)"g/mol O")="231.837 g/mol WO"_3"

$\text{H"_2"O} :$(2xx1.008color(white)(.)"g/mol H")+(1xx15.999"g/mol O")="18.015 g/mol H"_2"O"

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Mole Ratios
Determine the mol ratio between $\text{WO"_3}$ and $\text{H"_2"O}$ from the balanced equation.

$\left(1 \text{mol WO"_3)/(3"mol H"_2"O}\right)$ and $\left(3 {\text{mol H"_2"O")/(1"mol WO}}_{3}\right)$

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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"

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color(purple)("Moles Water"
Multiply mol ${\text{WO}}_{3}$ 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"

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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

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Theoretical Yield of Water is $\text{13.3 g}$.

Actual yield is $\text{10.0 g H"_2"O}$.

"Percent Yield"=("actual yield")/("theoretical yield")xx100

$\text{Percent Yield"=("10.0 g H"_2"O")/("13.3 g H"_2"O")xx100="75.2%}$ rounded to three significant figures