# Question 61803

Oct 20, 2015

${\text{2.35 g H"_3"PO}}_{4}$

#### Explanation:

Phosphorus pentachloride, ${\text{PCl}}_{5}$, will react with water to produce aqueous phosphoric acid, ${\text{H"_3"PO}}_{4}$, and hydrogen chloride vapor, $\text{HCl}$.

The balanced chemical equation for this reaction looks like this

${\text{PCl"_text(5(s]) + 4"H"_2"O"_text((l]) -> "H"_3"PO"_text(4(aq]) + color(red)(5)"HCl}}_{\textrm{\left(g\right]}}$

Notice that you have a $1 : 1$ mole ratio between phosphorus pentachloride and phosphoric acid, on one hand, and a $1 : \textcolor{red}{5}$ mole ratio between phosphorus pentachloride and hydrogen chloride, on the other.

You know for a fact that water will not be a limiting reagent, since you were told that it is in excess.

This means that all the moles of phosphorus pentachloride will react and produce phosphoric acid and hydrogen chloride according to the aforementioned mole ratios.

So, how many moles of phosphorus pentachloride you get in that many grams? Use the compund's molar mass to figure that out

5.00color(red)(cancel(color(black)("g"))) * ("1 mole PCl"_5)/(208.24color(red)(cancel(color(black)("g")))) = "0.02401 moles PCl"_5

Now, I'll show you how to find the number of grams of phosphoric acid produced by the reaction, and you can use the same technique to find the mass of hydrogen chloride.

So, how many moles of phosphoric acid will the reaction produce? Use the mole ratio that exists between the two compounds

0.02401color(red)(cancel(color(black)("moles PCl"_5))) * ("1 mole H"_3"PO"_4)/(1color(red)(cancel(color(black)("mole PCl"_5)))) = "0.02401 moles H"_3"PO"_4

To find how many grams would contain this many moles of phosphoric acid, use the compound's molar mass

0.02410color(red)(cancel(color(black)("moles"))) * "97.995 g"/(1color(red)(cancel(color(black)("mole")))) = color(green)("2.35 g H"_3"PO"_4)#

Now use the same method to find the grams of hydrogen chloride produced by the reaction.

Check out this cool video of how the reaction looks like

https://hml.cardiff.ac.uk/Play/255