Chemistry 2006/ 2007 Club***

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I am starting Chem II this semester. Anyone want to start a club??? :rolleyes:

only with transition metals

Please someone help me with stoichiometry and oxidation-reduction. I can't for the life of me understand this stuff and desperately need help.

An example of a stoichiometry problem I need help with is this:

Mg + H2SO4 ---> MgSo4 +H2

How do I figure out how many moles of H2 are produced by the complete reaction of 230 mg of Mg with sulfuric acid?

With oxidation-reduction, I can't figure out how to determine whether something is oxdized or reduced, for example:

CH4 (g) + 2 O2 (g) + H2O (g)

How do I figure out which is oxdized and which is reduced?

Please help. Thank you!!!!

first you must balance any reaction

mg + h2so4 ---> mgso4 +h2

well, you are lucky because it's already balanced. 1 is the stichiometry value for each. basically you are making sure what you started with has the same amount of atoms of each element on the reactant side that you ended with on the product side. check my post one or two pages back.

you have 230mg of mg. you know the molecular weight of mg by looking at the periodic table, which is 24.3 moles per gram. if you have 230mg, then you only have .23 grams, right? so if

(mass of reactant mg)/(molar mass of mg)= (moles mg) then

.23g/24.3g = 0.0095 moles of mg

then look at your balanced reaction

1 mg + 1 h2so4 ---> 1 mgso4 +1 h2

you have a 1 to 1 ratio of mg of reactant to h2 of product. so for every 0.0095 moles of mg you'll have 0.0095 moles of h2.

so then you do moles of h2 x molar mass of h2 = grams of h2

since you have 0.0095 moles of h2 (remember 1 to 1 ratio), you multiply that number by the molar mass of h2. well, you have two h molecules that have a molar mass of 1 each, so that means your equation would look like this

0.0095moles x 2(1g/mole)= grams of h2

or 0.019 grams of h2

it's very early in the morning and i'm sleepy so someone correct me if i'm wrong. time for bed.

thank you!!

Specializes in Telemetry Med/Surg.

With Oxidation and Reduction just remember that all of your answers are on the left side of the equation. Oxidation is the loss of electrons and is the atom or ion on the left side of the eqaution that becomes more positive as it moves to the right hand side of the equation. Also Oxidation is when the compound on the left hand side gains oxygen or loses hydrogen.

Reduction is the gain of electrons and is the atom or ion on the left side of the equation that becomes more negative as it moves to the right hand side of the equation. Also with Reduction, the compound on the left side of the equation loses oxygen or gains hydrogen. HINT::: O2 is always the oxidizing agent and H2 is always the reducing agent.

Therefore, O2 is going to be reduced and is your oxidizing agent and H2 is going to be oxidized and is your reducing agent.

oh yeah, forgot to answer that

i thought determining whether something is oxidized or reduced had something to do with the charge of the element or compound? when i had someone explain it to me they said that when a solid element turns to a liquid (aqueous) it is oxidized and vice versa, is this correct? thanks!

Specializes in Peds ER.
i thought determining whether something is oxidized or reduced had something to do with the charge of the element or compound? when i had someone explain it to me they said that when a solid element turns to a liquid (aqueous) it is oxidized and vice versa, is this correct? thanks!

It is my understanding, and this is just a newbie at this, oxidation and reduction are associated with gain or loss of electrons and nothing else.

I can think of examples in which something might lose an electron, thus oxidizing, and not liquify. Case in point, iron oxidizes. Iron oxide is still a solid.

I remember that REDUCTION is called that because the charge is REDUCED or lessened. I originally thought "Well if you GAIN an electron why do you call it REDUCTION" but it makes sense now. CHARGE is reduced.

Once I remember that, I remember Oxidation means increase in charge. Increase in charge is only possible through loss of electrons.

Yep.

please can anyone help me with the following q's

Write formulas for conjugate base of HCO3^-

write formulas for the conjugate acid of PO4^ 3-

predict whether each of the following pair of susbtance could function as a buffer system in aqueous solution HCN and KCN

thanks

reactions look like this

HCl + KOH ---> KCl + H2O

you can replace them with this terminology or flip it backwards

acid + base ---> conjugate base + Conjugate acid

Very simple... the acid, once it loses a Hydrogen (for simplicity, acids always lose an H) becomes a conjugate base. Most equations in Gen chem are substitution reactions, so you are just swapping atoms. The same with the base... the base, once it gains a H (taken from the acid) becomes a conjugated acid (notice H2O has an extra H... it was KOH before and only had one H).

If your acid is a STRONG acid then your conjugate will be a weak base. If your acid was weak, the conjugate base would be strong, etc. Same with bases, if your base you started with was strong, your conjugate acid would be weak.

So, in your case, HCO3 and PO4 would be

HCO3 (HCO3 is tricky because it could be an acid AND or base in some reactions)

HCO3 conjugate base is CO3

HCO3 conjugate acid is H2CO3

PO4 (base) conjugate acid is HPO4

Remember, in order to obtain a conjugate base, you look at the appropriate conjugate acid and remove one (and only one) H+ ion from it. Be sure to remember that acid-base conjugate pairs differ by a single H+ ion.

Sometimes they can be tricky, but just remember the rule

H3O + OH ----> H2O + H2O

acid + base ----> ca + cb

or flip the one above

HCl + KOH ---> KCl + H2O

flipped

KCl + H2O ---> HCl + KOH

base + acid ---> Conjugate acid + Conjugate base

just remember the rule

predict whether each of the following pair of susbtance could function as a buffer system in aqueous solution HCN and KCN

thanks

I can help you if you don't post the "following substances"

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