Question #6308e

1 Answer
Aug 21, 2017

Answer:

#4.3# #"kg H"_2"O"#

Explanation:

We're asked to find the mass (in #"kg"#) of water produced from the combustion of #3.8# #"L C"_8"H"_18#, given its density.

Let's first write the balanced chemical equation for this reaction:

#ul(2"C"_8"H"_18(l) + 25"O"_2(g) rarr 16"CO"_2(g) + 18"H"_2"O"(g)#

We'll convert from liters of #"C"_8"H"_18# to grams using its density:

#3.8cancel("L C"_8"H"_18)((10^3cancel("mL"))/(1cancel("L")))((0.79color(white)(l)"g C"_8"H"_18)/(1cancel("mL C"_8"H"_18))) = color(red)(ul(3.0xx10^3color(white)(l)"g C"_8"H"_18#

Now, we'll use the molar mass of octane (#114.229# #"g/mol"#) to convert from grams to moles:

#color(red)(3.00xx10^3)cancel(color(red)("g C"_8"H"_18))((1color(white)(l)"mol C"_8"H"_18)/(114.229cancel("g C"_8"H"_18))) = color(green)(ul(26.3color(white)(l)"mol C"_8"H"_18#

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of water that form:

#color(green)(26.3)cancel(color(green)("mol C"_8"H"_18))((18color(white)(l)"mol H"_2"O")/(2cancel("mol C"_8"H"_18))) = color(purple)(ul(236color(white)(l)"mol H"_2"O"#

Lastly, we'll use the molar mass of water (#18.015# #"g/mol"#) to find its mass:

#color(purple)(236)cancel(color(purple)("mol H"_2"O"))((18.015cancel("g H"_2"O"))/(1cancel("mol H"_2"O")))((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(blue)(ulbar(|stackrel(" ")(" "4.3color(white)(l)"kg H"_2"O"" ")|)#

rounded to #2# significant figures.