Lisa wants to make an acid. She has two possible compounds she could add to 250.0 g of water to make this acid – EITHER 234.5 g of calcium oxide OR 678.9 g of tetraphosphorus decaoxide (she won’t use both, just one).? a. Which compound should s

Lisa wants to make an acid. She has two possible compounds she could add to 250.0 g of water to
make this acid – EITHER 234.5 g of calcium oxide OR 678.9 g of tetraphosphorus decaoxide (she won’t use both, just one).

a. Which compound should she use? Explain your answer (with words and full sentences), and write any balanced chemical reactions that would help you justify your claim.

b.
What is the chemical formula of the acid, and what is the maximum mass of the acid she could theoretically produce when she does this reaction?

c. What is the EXCESS reactant, and how many grams of this reactant are leftover when the reaction is completed?

1 Answer
Nov 8, 2015

Answer:

Here's what I got.

Explanation:

So, you know that Lisa must mix either calcoium oxide, #"CaO"#, or phosphorus pentoxide, #"P"_4"O"_10#, with water in order to produce an acid.

SIDE NOTE Phosphorus pentoxide is just the common name used for tetraphosphorus decoxide. The rationale for this naming convention is that it's actually the name of tetraphosphorus decoxide's empirical formula, which is #"P"_2"O"_5#.

Now, calcium oxide will react with water to form calcium hydroxide, #"Ca"("OH")_2#, which is considered a strong base.

#"CaO"_text((s]) + "H"_2"O"_text((l]) -> "Ca"("OH")_text(2(s])#

Calcium hydroxide will actually precipitate out of solution. Mind you, calcium hydroxide (also called slacked lime) is still considered a strong base because, despite the fact that it's not very soluble in aqueous solution, what dissolves will dissociate completely to form

#"Ca"("OH")_text(2(aq]) -> "Ca"_text((aq])^(2+) + 2"OH"_text((aq])^(-)#

It's obvious that LIsa should not use calcium oxide for her reaction, since it reacts with water to form a base.

On the other hand, phosphorus pentoxide will react with water to form phosphoric acid, a triprotic weak acid.

#"P"_4"O"_text(10(s]) + color(red)(6)"H"_2"O"_text((l]) -> color(blue)(4)"H"_3"PO"_text(4(aq])#

So there you have it, Lisa must use phosphorus pentoxide to form her acid solution.

Now, in order to determine what the maximum amount of acid the reaction can produce, you need to take a look at the mole ratios that exist between the species that take part in the reaction.

Notice that #1# mole of #"P"_4"O"_10# must react with #color(red)(6)# moles of water to produce #color(blue)(6)# moles of phosphoric acid.

To determine how many moles of phosphorus pentoxide and how many moles of water Lisa has in her samples, use the two compounds' respective molar masses

#678.9color(red)(cancel(color(black)("g"))) * ("1 mole P"_4"O"_10)/(283.89color(red)(cancel(color(black)("g")))) = "2.3914 moles P"_4"O"_10#

and

#250.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "13.877 moles H"_2"O"#

And now the question is - do you have enough moles of water to allow for all the moles of phosphorus pentoxide to react?

This many moles of phosphorus pentoxide would require

#2.3914color(red)(cancel(color(black)("moles P"_4"O"_10))) * (color(red)(6)" moles H"_2"O")/(1color(red)(cancel(color(black)("mole P"_4"O"_10)))) = "14.348 moles H"_2"O"#

Since you don't have enough moles of water present, the water will act as a limiting reagent. In other words, phosphorus pentoxide will be the excess reagent.

This means that the reaction will only consume

#13.877color(red)(cancel(color(black)("moles H"_2"O"))) * ("1 mole P"_4"O"_10)/(color(red)(6)color(red)(cancel(color(black)("moles H"_2"O")))) = "2.3128 moles P"_4"O"_10#

You will thus be left with

#n_"remaining" = 2.3914 - 2.3128 = "0.07860 moles P"_4"O"_10#

In grams, this is equivalent to

#0.07860color(red)(cancel(color(black)("g"))) * "283.89 g"/(1color(red)(cancel(color(black)("mol")))) = color(green)("22.31 g P"_4"O"_10)#

The reaction will produce

#2.3128color(red)(cancel(color(black)("moles P"_4"O"_10))) * (color(blue)(4)" moles H"_3"PO"_4)/(1color(red)(cancel(color(black)("mole P"_4"O"_10)))) = "9.2512 moles H"_2"O"#

In grams, this is equivalent to

#9.2512color(red)(cancel(color(black)("moles"))) * "97.995 g"/(1color(red)(cancel(color(black)("mol")))) = color(green)("906.2 g H"_3"PO"_4)#