Question #333ff

1 Answer
Aug 3, 2017

Answer:

#32.1# #"g H"_2"O"#

Explanation:

We're asked to find the volume, in liters, of #"H"_2"O"(g)# that form when #100# #"g CaO"# react.

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

#3"CaO"(aq) + 2"H"_3"PO"_4(aq) rarr "Ca"_3"(PO"_4")"_2(aq) + 3"H"_2"O"(g)#

Let's use the molar mass of #"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 #"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 #"H"_2"O"# to find the number of grams that can form:

#color(green)(1.78)cancel(color(green)("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"" ")|)#