Question

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:

`2H_2(g) + O_2(g) -> 2H_2O(L)`

Answer 1

WARNING! Long Answer! The theoretical yield of water is 10.1 g.
`"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 `"H"_2`

From the Ideal Gas Law,

`n = (PV)/(RT)`

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

`V = "13.74 L"`

`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`

`T = "(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 `"O"_2`

`P = "801 torr" = "1.054 atm"`
`V = "6.53 L"`
`R = "0.082 06 L·atm·K"^"-1""mol"^"-1"`
`T = "(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 `"H"_2"O"` formed from the `"H"_2`

`"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 `"H"_2"O"` formed from the `"O"_2`

`"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 `"O"_2`, because it produces the fewest moles of product.

Calculate the theoretical yield of `"H"_2"O"`

`"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 %`