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 6.55 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(g) -> H_2O (l)`

Answer 1

Theoretical yield: `"10.5` `"g H"_2"O"`

Limiting reactant: `"H"_2`

Percent yield: `82.9%`

Explanation:

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

`2"H"_2(g) + "O"_2(g) rarr color(red)(2)"H"_2"O"(g)`

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:

`PV = nRT`

The units for pressure must be in `"atm"`, so let's convert it:

`P = 801cancel("torr")((1color(white)(l)"atm")/(760cancel("torr"))) = 1.05` `"atm"`

The temperature must also be in `"Kelvin, K"`:

`T_ ("H"_2) = 30.0^"o""C" + 273 = 303` `"K"`

`T_ ("O"_2) = 25^"o""C" + 273 = 298` `"K"`

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

`n = (PV)/(RT)`

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:

`"H"_2`: `(0.582color(white)(l)"mol")/(2 "(coefficient)") = 0.291`

`"O"_2`: `0.156` (coefficient of `1`)

Thus, hydrogen is the limiting reactant, so we'll use the mole values of `"H"_2` for our calculations.

To find the number of grams of `"H"_2"O"` that are able to form, let's first find the relative number of moles of `"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` `"mol H"_2"O"`

To calculate this value in grams, we'll use its molar mass (`18.02` `"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"_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%`