Question #1e405

1 Answer
Aug 29, 2016

Answer:

Here's what I got.

Explanation:

You said that you understand the first part, so I'll just focus on the second part here.

So, the balanced chemical reaction looks like this

#"NH"_ 4"Cl"_ ((aq)) + "NaOH"_ ((aq)) -> "NH"_ (3(aq)) + "H"_ 2"O"_ ((l)) + "NaCl"_ ((aq))#

The chemical equation tells you that you every mole of ammonium chloride, #"Nh"_4"Cl"#, that takes part in the reaction consumes #1# mole of sodium hydroxide, #"NaOH"#.

Moreover, every mole of ammonium chloride that takes part in the reaction produces #1# mole of ammonia, #"NH"_3#, #1# mole of water, and #1# mole of sodium chloride, #"NaCl"#.

This essentially means that if you know how many moles of a reactant take part in the reaction, you know how many moles of the other chemical species will be involved.

Now, the calculations show that you're mixing

#"0.50 moles NH"_4"Cl "# and #" 0.25 moles NaOH"#

The question to answer now is

Do you have enough moles of each reactant to allow both of them to be completely consumed?

If not, one of them will act as a limiting reagent, which is a fancy way of saying that it will run out before all the moles of the second reactant get the chance to react.

In this case, sodium hydroxide will be the limiting reagent because you have fewer moles of this reactant than of ammonium chloride.

In other words, #0.50# moles of ammonium chloride would require

#0.50 color(red)(cancel(color(black)("moles NH"_4"Cl"))) * "1 mole NaOH"/(1color(red)(cancel(color(black)("mole NH"_4"Cl"))))#

# = " 0.50 moles NaOH"#

to react completely. But you only have #0.25# moles of sodium hydroxide available, so the sodium hydroxide will be completely consumed, along with

#0.25 color(red)(cancel(color(black)("moles NaoH"))) * ("1 mole NH"_4"Cl")/(1color(red)(cancel(color(black)("mole NaoH"))))#

# = " 0.25 moles NH"_4"Cl"#

So, the reaction consumes #0.25# moles of each reactant, which means that you'll be left with

#overbrace("0.25 moles NaOH")^(color(blue)("what you start with")) - overbrace("0.25 moles NaOH")^(color(purple)("what gets consumed")) #

#= " 0 moles NaOH " -># completely consumed

and

#overbrace("0.50 moles NNH"_4"Cl")^(color(blue)("what you start with")) - overbrace("0.25 moles NH"_4"Cl")^(color(purple)("what gets consumed"))#

# = " 0.25 moles NH"_4"Cl " -># in excess

So, to answer your questions

(a) You can compare the ratio of the reactants with ammonia and not with the other products because they are all produced in #1:1# mole ratios.

Since the reaction produces equal numbers of moles of ammonia, water, and sodium chloride, there's no point in comparing the numbers of moles of reactants with all three products.

You can do that, but all you really need to do is pick one product to compare the reactants with.

(b) I hope that I have answered this one above. You subtract the number of moles of sodium hydroxide from the number of moles of ammonium chloride because the former is present in the lesser amount.

Your goal is to find the number of moles of the reactant that is in excess, so always look to subtract the number of moles of the limiting reactant from the number of moles of the excess reactant.