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)#

1 Answer
Jun 24, 2017

Answer:

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%#