# 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: $2 {H}_{2} \left(g\right) + {O}_{2} \left(g\right) \to 2 {H}_{2} O \left(L\right)$

Sep 15, 2016

WARNING! Long Answer! The theoretical yield of water is 10.1 g.
${\text{O}}_{2}$ 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 ${\text{H}}_{2}$

From the Ideal Gas Law,

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

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

$V = \text{13.74 L}$

$R = \text{0.082 06 L·atm·K"^"-1""mol"^"-1}$

$T = \text{(30.0 + 273.15) K" = "303.15 K}$

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 ${\text{O}}_{2}$

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

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 $\text{H"_2"O}$ formed from the ${\text{H}}_{2}$

$\text{Moles of H"_2"O" = 0.5821 color(red)(cancel(color(black)("mol H"_2))) × ("2 mol H"_2"O")/(2 color(red)(cancel(color(black)("mol H"_2)))) = "0.5821 mol H"_2"O}$

b. Calculate the moles of $\text{H"_2"O}$ formed from the ${\text{O}}_{2}$

$\text{Moles of H"_2"O" = 0.2813 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol H"_2"O")/(1 color(red)(cancel(color(black)("mol O"_2)))) = "0.5626 mol H"_2"O}$

The limiting reactant is ${\text{O}}_{2}$, because it produces the fewest moles of product.

Calculate the theoretical yield of $\text{H"_2"O}$

$\text{Theoretical yield" = 0.5626 color(red)(cancel(color(black)("mol H"_2"O"))) × ("18.02 g H"_2"O")/(1 color(red)(cancel(color(black)("mol H"_2"O")))) = "10.1 g H"_2"O}$

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 %