Al powder and NaOH

[luken93]

Question 5
One method to determine the empirical formula of a compound of lead and chlorine is to dissolve the solid
in hot water and then add aluminium powder. A chemical reaction occurs that causes solid lead metal to
form and aluminium ions to be produced. The results from such an experiment are shown below.
mass of lead chloride compound 1.213 g
mass of aluminium powder added 3.384 g
mass of lead metal recovered 0.902 g
a. Some of the aluminium powder remained unreacted. The unreacted aluminium powder was dissolved
using sodium hydroxide solution.
Explain why this step was necessary.

I have no idea about this, or am I missing something?

[shinny]

Well since you're not sure about the empirical formula of the lead and chlorine compound, you won't be able to do the stoic directly. Instead, you'll have to react what you do know (the aluminium powder) with something else you know (NaOH solution). From there, you can work out how much aluminium didn't react, and how much did react, and hence calculate the empirical formula.

[luken93]

Well since you're not sure about the empirical formula of the lead and chlorine compound, you won't be able to do the stoic directly. Instead, you'll have to react what you do know (the aluminium powder) with something else you know (NaOH solution). From there, you can work out how much aluminium didn't react, and how much did react, and hence calculate the empirical formula.

I worked out the empirical of Lead (II) Chloride to be PbCl₂, but I don't understand why the step of dissolving the unreacted aluminium powder is necessary.
It's necessary for what?

[shinny]

Oh sorry my previous answer was total rubbish; haven't done this stuff in too long. I think it's to remove all the remaining aluminum powder so that a gravimetric analysis can be performed of the lead solid that remains. Right now there's not really any method of isolating it.

[luken93]

Oh sorry my previous answer was total rubbish; haven't done this stuff in too long. I think it's to remove all the remaining aluminum powder so that a gravimetric analysis can be performed of the lead solid that remains. Right now there's not really any method of isolating it.

ummm ok i think that its a bit over my head for yr 11, i will ask my teacher today

[luken93]

The correct answer straight from NEAP 2010:

a. Unused aluminium powder must be dissolved, otherwise, when the lead metal is compacted
and removed for weighing, some of the aluminium will be trapped and produce an inaccurate
mass of lead.

A rather silly question come to think of it, trying to make it harder than it is

[luken93]

Okay another question, which of the following is incorrect:
A. radius of an aluminium atom > radius of an aluminium ion
B. electronegativity of chlorine > electronegativity of bromine
C. radius of an oxygen atom < radius of an oxygen ion
D. number of valence electrons in a magnesium atom > number of valence electrons in a magnesium ion

I would've said C, because the more electrons in the outer shell, the tighter they are pulled together, hence radius of oxygen atom would be greater that oxygen ion, but the answer is D?

[Martoman]

D is defs incorrect. Magnesium ion it gives away electrons.

It thus has LESS electrons than its atom.

[luken93]

D is defs incorrect. Magnesium ion it gives away electrons.

It thus has LESS electrons than its atom.

but doesnt it then have a full outer shell of 8 valence electrons, because it takes the form of a noble gas?

[Martoman]

Oh I see what you mean here. yes, but i think the question writer simply "meant" what I said and wrote it wrong.

[luken93]

Yes, Martoman is probably right. That's the flaw in the question. (I think they meant to exclude the word valence)
but anyway... think about it that way. You initially picked C as the answer, but if C was right, A would have to be right too

No it wouldn't because A has a different number of shells than the other, because it drops back to the previous period