# Question 333ff

Aug 3, 2017

$32.1$ $\text{g H"_2"O}$

#### Explanation:

We're asked to find the volume, in liters, of $\text{H"_2"O} \left(g\right)$ that form when $100$ $\text{g CaO}$ react.

First, let's write the balanced chemical equation for this reaction:

$3 \text{CaO"(aq) + 2"H"_3"PO"_4(aq) rarr "Ca"_3"(PO"_4")"_2(aq) + 3"H"_2"O} \left(g\right)$

Let's use the molar mass of $\text{CaO}$ to find the number of moles that react:

100cancel("g CaO")((1color(white)(l)"mol CaO")/(56.077cancel("g CaO"))) = color(red)(ul(1.78color(white)(l)"mol CaO"

Now, we'll use the coefficients of the chemical equation to find the relative number of moles of $\text{H"_2"O}$ that can form:

color(red)(1.78)cancel(color(red)("mol CaO"))((3color(white)(l)"mol H"_2"O")/(3cancel("mol CaO"))) = color(green)(ul(1.78color(white)(l)"mol H"_2"O"#

Finally, we'll use the molar mass of $\text{H"_2"O}$ to find the number of grams that can form:

$\textcolor{g r e e n}{1.78} \cancel{\textcolor{g r e e n}{\text{mol H"_2"O"))((18.015color(white)(l)"g H"_2"O")/(1cancel("mol H"_2"O"))) = color(blue)(ulbar(|stackrel(" ")(" "32.1color(white)(l)"g H"_2"O"" }} |}$