# Question 6308e

Aug 21, 2017

$4.3$ $\text{kg H"_2"O}$

#### Explanation:

We're asked to find the mass (in $\text{kg}$) of water produced from the combustion of $3.8$ ${\text{L C"_8"H}}_{18}$, 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 ${\text{C"_8"H}}_{18}$ 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 ($114.229$ $\text{g/mol}$) 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 ($18.015$ $\text{g/mol}$) to find its mass:

$\textcolor{p u r p \le}{236} \cancel{\textcolor{p u r p \le}{\text{mol H"_2"O"))((18.015cancel("g H"_2"O"))/(1cancel("mol H"_2"O")))((1color(white)(l)"kg H"_2"O")/(10^3cancel("g H"_2"O"))) = color(blue)(ulbar(|stackrel(" ")(" "4.3color(white)(l)"kg H"_2"O"" }} |}$

rounded to $2$ significant figures.