andrew's really crappy chem thread

[soNasty]

hey everyone, currently trying my best to cement all knowledge from last year to prepare for this year
i'll be asking all my chem-based questions in this thread
this is my weakest subject by far - trying hard though, to make it in my top 4!

ok so:

how would i write an equation for a question such as:

1. The iodide ions in a solution containing 0.300g of sodium iodide were precipitated as silver iodide. What mass of silver iodide was formed?

2. A 0.500 g sample of sodium sulfate (Na2SO4) and a 0.500 g sample of aluminium sulfate (Al2(SO4)3) were dissolved in a volume of water, and excess barium chloride was added to precipitate barium sulfate.
What was the total mass of barium sulfate produced?

Also, does anyone have any tips for doing these sort of questions?
The wording is simple however i'm almost always unsure as to the order of the reaction (eg where reactants and products go)
calculating everything (number wise) is fine.

Thank you all

[MagicGecko]

1. It says that sodium iodide (NaI) was precipitated (caused the production of) as silver iodide (AgI). This means that sodium iodide is the reactant and silver iodide is the product. In chemistry, the reactants are always on the left and the products are always on the right so you would end up with something like this:

But where did the silver come from? The question doesn't point it down directly, but since sodium iodide produced silver iodide, this means that there was silver in the solution, meaning that this will be the equation:


2. With this question, you would have 3 different ionic questions because they don't give any info about barium or chloride so the only way you can find the mass of barium sulfate is to find the amount of SO4^2- since that is what they have in common.
so, net ionic equations:


         and the one for the product

Also, does anyone have any tips for doing these sort of questions?

Its always good to know your solubility table so you know how to assign states to the reactants and products. Also make sure you know what each of their charges are (Barium has a charge of 2+ for example). Don't worry, once you do a ton of questions, equation writing on the exam will come really easy

[soNasty]

thank you

[soNasty]

can anyone help me with identifying appropriate indicators (methyl orange and phenolphthalein) for this question please? need to cement this info properly.

[brightsky]

There is a table in the data booklet that gives you the pH ranges of different indicators. Choose the indicator which minimizes the titration error (i.e. the difference between the end point and the equivalence point).

[soNasty]

thank you brightsky!
however it only gives you 2 indicators to choose from already!

[soNasty]

How would I go about in setting up a reaction for this question?

A 1.20 g antacid tablet contains 80.0% by mass of magnesium hydroxide as the active ingredient. What volume of 0.1500 M hydrochloric acid could the antacid tablet neutralise?

[brightsky]

2HCl(aq) + Mg(OH)2(s) --> MgCl2(aq) + 2H2O(l)

EDIT: Woops, magnesium hydroxide is highly insoluble in water.

[soNasty]

What would the products of a reaction between Sodium hydroxide and ammonium consist of, if any?

The question corresponding to this is:
 Lawn fertiliser contains ammonium ions (NH4+). A 1.234 g sample of lawn fertiliser was dissolved in water to make a 250.0 mL solution. A 20.00 mL aliquot of this solution was added to a flask containing 20.00 mL of 0.1022 M sodium hydroxide solution. The flask was heated until the reaction:
NH4+(aq) + OH–(aq) ® NH3(aq) + H2O(l)
was complete. Excess sodium hydroxide in the resulting solution was titrated with 0.1132 M hydrochloric acid, using phenolphthalein as indicator. The end point was reached when 9.97 mL had been added.

[scribble]

You should be able to recognise that sodium hydroxide is a strong base and that ammonium is a weak acid.
When you put them together, the acid and the base are neutralised, creating water, and you're left with ammonia. the sodium is a spectator ion.

[MagicGecko]

Just as sribble said, the Na acts as a spectator ion so that's why you have OH- instead of NaOH.
NH4+ acts as an acid and donates its proton becoming NH3 whereas OH- acts as a base and accepts that proton therefore becoming H20

[soNasty]

Thank you, yeah I get that.
What would the reaction look like with the spectator ions?

[emilyhobbes]

It hasn't given you the other "half" of the ammonium compound, so you can't really write a full equation, but from the info given it'd look like
NaOH(aq) + NH4+(aq) -> NH3(aq) + H2O (l) + Na+(aq)

[soNasty]

It hasn't given you the other "half" of the ammonium compound, so you can't really write a full equation, but from the info given it'd look like
NaOH(aq) + NH4+(aq) -> NH3(aq) + H2O (l) + Na+(aq)

Ah okay thanks a lot!

[soNasty]

how would i show what has been oxidised and reduced in:



with and without the use of oxidation numbers? since this is a question from the textbook before the explanation of oxidation numbers in redox reactions..