# 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 6.55 L of oxygen gas at 25 C and 801 torr are drawn into a cylinder where the following reaction takes place. $2 {H}_{2} + {O}_{2} \left(g\right) \to {H}_{2} O \left(l\right)$

Jun 24, 2017

Theoretical yield: "10.5 $\text{g H"_2"O}$

Limiting reactant: ${\text{H}}_{2}$

Percent yield: 82.9%

#### Explanation:

Note: this equation is not fully balanced; here is what it should look like:

$2 \text{H"_2(g) + "O"_2(g) rarr color(red)(2)"H"_2"O} \left(g\right)$

To solve this equation, let's first find the limiting reactant, by using the ideal-gas equation to find the moles of each reactant present:

$P V = n R T$

The units for pressure must be in $\text{atm}$, so let's convert it:

P = 801cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 1.05 $\text{atm}$

The temperature must also be in $\text{Kelvin, K}$:

T_ ("H"_2) = 30.0^"o""C" + 273 = 303 $\text{K}$

T_ ("O"_2) = 25^"o""C" + 273 = 298 $\text{K}$

Let's now rearrange the ideal-gas equation to solve for the number of moles, $n$:

$n = \frac{P V}{R T}$

and plug in the values for each reactant to find their quantity:

n_ ("H"_2) = ((1.05cancel("atm"))(13.74cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(303cancel("K"))) = color(red)(0.582 color(red)("mol H"_2

n_ ("O"_2) = ((1.05cancel("atm"))(6.55cancel("L")))/((0.082057(cancel("L")·cancel("atm"))/("mol"·cancel("K")))(298cancel("K"))) = color(blue)(0.156 color(blue)("mol O"_2

We find the limiting reactant by dividing each by the coefficient in front of it in the chemical equation; the lower value is the limiting reactant:

${\text{H}}_{2}$: $\left(0.582 \textcolor{w h i t e}{l} \text{mol")/(2 "(coefficient)}\right) = 0.291$

${\text{O}}_{2}$: $0.156$ (coefficient of $1$)

Thus, hydrogen is the limiting reactant, so we'll use the mole values of ${\text{H}}_{2}$ for our calculations.

To find the number of grams of $\text{H"_2"O}$ that are able to form, let's first find the relative number of moles of $\text{H"_2"O}$ using the coefficients of the equation:

color(red)(0.582)cancel(color(red)("mol H"_2))((2color(white)(l)"mol H"_2"O")/(2cancel("mol H"_2))) = 0.582 $\text{mol H"_2"O}$

To calculate this value in grams, we'll use its molar mass ($18.02$ $\text{g/mol}$):

0.582cancel("mol H"_2"O")((18.02color(white)(l)"g H"_2"O")/(1cancel("mol H"_2"O"))) = color(blue)(10.5 color(blue)("g H"_2"O"

If $8.7$ "$g H \text{_2"O}$ formed, then the percent yield of water for this reaction is

%"yield H"_2"O" = "actual yield"/"theoretical yield"xx100%

= (8.7cancel("g H"_2"O"))/(color(blue)(10.5)cancel(color(blue)("g H"_2"O"))) xx 100% = color(green)(82.9%