Chemistry Tutorials

[Raiser Blade]

Disclaimer: I am not looking for straight answers. I am looking for a little help understanding the following concepts. My regular chemistry teacher is having a baby and our sub did a less than excellent job explaining these kind of problems.

Here are a few practice problems that I just can't seem to figure out what to do.


If a chemist combines 20.0 ml of a 5.5 M solution of H3PO4 with excess RbOH
How many grams of Rb3PO4 will be produced?

From what I could pick up I think I have to balance the equation. After that it gets a little fuzzy. There was also something about H3PO4 being the limiting reactant or something.


The next thing was about titration.

A chemist performs a titration of C2H4O2 and NH3. If it took 86 ml of 5.0 M solution of NH3 to neutralize 250 ml of C2H4O2 what is the molarity of
C2H4O2?


If you can give me a demonstraton of how to work these problems then great. Heck if you can just tell me what to do in what order than that would be cool too. So thanks in advance :smallsmile:

[Ryshan Ynrith]

Right. For the first one, you first need to balance the equation between the two compounds. Once that is done, convert the volume of reagent you have into moles via its molarity (moles/liter), use the stoichiometric ratio of the equation to determine the number of moles of your product, and then multiply by the molecular weight of your product to determine grams:

Mols/L*L=mols, mols (A)->mols (B), mols*g/mols=g. This is referred to as stoichiometry. You can also (if provided a weight in the beginning) begin by dividing by the atomic mass to find moles of A.

The second is similar; recognize that molarity is moles/liter and find the number of moles of your NH_3, balance the equation, and divide the number of moles of your other compound by the volume to determine molarity.

Be sure always to carry your units through-you can be sure you have the right answer if your units make physical sense. So if you're writing down the math, put the units next to the numbers, and always convert to standard measurements (liters for volume, meters for length, and so forth) unless you think you can keep track of the conversions in your head.

So say you have 57 ml of 7 M Na (this makes no actual sense, but it's an example) reacting with excess Cl solution, and you want to find the number of moles of Cl you are reacting with. You have 7*.057=.371 mols Cl, the ratio of Cl to Na in reaction is 1, so you have .371*1=.371 moles of Na reacting.

If you titrate 250 ml of Cl with the 57 ml of 7 M Na, that means that there were only .371 mols of Cl in the 250 ml, which means that the Cl solution is .371/.250=1.484 M.

Hope this helps!

[Solo]

Disclaimer: I am not looking for straight answers. I am looking for a little help understanding the following concepts. My regular chemistry teacher is having a baby and our sub did a less than excellent job explaining these kind of problems.

Here are a few practice problems that I just can't seem to figure out what to do.


If a chemist combines 20.0 ml of a 5.5 M solution of H3PO4 with excess RbOH
How many grams of Rb3PO4 will be produced?

From what I could pick up I think I have to balance the equation. After that it gets a little fuzzy. There was also something about H3PO4 being the limiting reactant or something.

You will run out of H3PO4 first, so find the mols of H3PO4 you have, then find out how many mols of RbOH you can make with it via the ratio of mols of PO in the compound Rb3PO4.

You will find that in the final product, 3 atoms of Rb combines with 3 molecule of PO4. Therefore the ratio between these two components is 3:1. As we have an excess of Rb, you need not worry about it.

Anyways, you'll be able to form an amount of 3PO4 equal to three times the mols of HPO4 you have. That would be 5.5M*.0200L=0.11 mol H3PO4, so you get 0.33 mols of your final product.



The next thing was about titration.

A chemist performs a titration of C2H4O2 and NH3. If it took 86 ml of 5.0 M solution of NH3 to neutralize 250 ml of C2H4O2 what is the molarity of
C2H4O2?



Find out how many mols of NH3 were used to neutralize C2H4O2, then find out how many mols it takes to neutralize C2H4O2. You will now be able to find the amount of mols of C2H4O2 present in 250ml of C2H4O2 solution.

Let's see... C2H4O2 is acetic acid (although it could be something else with the same formula, an isomer). I'm not sure how it interacts with ammonia, but I know that nitrogen has a lone electron pair, as it cannot form double bonds with Hydrogen. Therefore, this lone Electron pair will....?

I'm going to guess and say it takes a hydrogen from acetic acid, neutralizing the ammonia, and turning it into ammonium (which is a weak acid, I believe)

This means one mol of acetate will neutralize one mol of NH3.

We have 5.0M*0.086L= 0.43 mols of Acetic acid. This leads me to believe that we will be able to neutralize 0.43 mols of NH3. Divide by 0.250L to get 1.72 M Acetic Acid.

I did this after just getting up, so I can't guarantee if it is right or not.

[Raiser Blade]

I'm pretty sure I grasp the concept. Thank you guys so much! :smallsmile:

[Raiser Blade]

Okay new week new concept. This problem stumped me. I can't even understand how to go about solving it.

Here is the problem.

Use this equation to solve the following: 3K2CO3 + 2Al(NO3)3 = Al2(CO3)3 + 6 KNO3

A chemist needs 350.0 grams of Al2(CO3)3, How many milliliters of a 3.5 M solution of K2CO3 must be added to excess Al(NO3)3?

[RandomLogic]

First you want to convert everything to one set of units. So convert the grams of AL2(CO3)3 to moles.

So then you work that through the balanced chemical reaction to figure out how many moles of K2CO3 you need. Since its a 3 to 1 ratio its 3 times the mols of AL2(co3)3.

Now you are given a 3.5 M solution of K2CO3 and what you do here is find how many mili-liters you need to get that many moles. (M/L=mols)

I think that should do it.

[Solo]

Okay new week new concept. This problem stumped me. I can't even understand how to go about solving it.

Here is the problem.

Use this equation to solve the following: 3K2CO3 + 2Al(NO3)3 = Al2(CO3)3 + 6 KNO3

A chemist needs 350.0 grams of Al2(CO3)3, How many milliliters of a 3.5 M solution of K2CO3 must be added to excess Al(NO3)3?

You must visualize the concepts presented! They are simply asking you to find the number of moles of Al(NO3)3, and figure out how many moles of K2CO3 must be added to that in order to get the number of moles of 350.0 grams of Al2(CO3)3

[Raiser Blade]

Once again thank you. :smallsmile: