PDA

View Full Version : Help with High School Chemistry. Oh.


Cat's Pajamas
Feb 6th, 2007, 02:10 AM
Please :sobbing:

I do not get this Stoichiometry at all :help:

I think that if someone could help me with a few than I could do the rest, but for now I'm just completely :weirdo:-ed

Here's a few problems, and I would request that you show me how to do it and not just the answer. I've seen some Chemistry wizards on this board before. ;)

---------

In problems involving reduction with carbon, assume the carbon is completely oxidized to carbon dioxide. (Um whatever that means :o).

Mass-Mass Problems

1. Carbon dioxide is produced in the reaction between calcium carbonate and hydrochloric acid. How many grams of calcium carbonate would be needed to react completely with 15.0 grams of hydrochloric acid? How many grams of carbon dioxide would be produced in this reaction? How many grams of calcium chloride would be formed at the same time?

2. Sulfur dioxide may be cartlytically oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process of 100.0 g of oxygen are available for the oxidation?

That's all for now :o

LeRoy.
Feb 6th, 2007, 02:16 AM
1)
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

1 MOL + 2 MOL --> 1 MOL + 1 MOL + 1 MOL

FW

CaCO3 = 100
HCl = 36.5
CO2 = 44
CaCl2 = 111

so 100 gms of CaCO3 would react with 2*36.5 gms of HCl to produce 111 gms of CaCl2 and 44 gms of CO2

For 15 gms do the calculations

Brooklyn90
Feb 6th, 2007, 02:16 AM
omg we just took a test over Stoichiometry today in my chemisty class. :lol: the only thing is I have no idea what im fucking doing :o I have like a D in the calss :lol: :p
Hopfully somone can help you :wavey:

LeRoy.
Feb 6th, 2007, 02:19 AM
Sulfur dioxide may be cartlytically oxidized to sulfur trioxide. How many grams of sulfur dioxide could be converted by this process of 25.0 g of oxygen are available for the oxidation?

SO2 + 1/2 O2 --> SO3

1 mol + 1/2 mol --> 1 mol

FW

SO2 = 64
O2 = 32
SO3 = 80

64 gms of SO2 would react with 16 gms of O2 to give 80 gms of SO3

Calculate for 25 gms of O2 using direct proportion

angele87
Feb 6th, 2007, 02:25 AM
The first thing you need to do with any problem like this is figure out the complete, balanced equation. For the first one , we have:

CaCO3 + 2HCl ----> C02 + CaCl2 + H20.

With these kinds of reactions, figuring our the formula is a little tricky. Usually though when you have C03 turned into C02, water is also a product. Now you know that 1 mol of CaCO3 and 2 mols of HCl make 1 mol of C02.

In the problem, you're given the weight of the hydrochloric acid, which is 15.0 grams. You need to convert that into mols. To do that, find the molecular mass ( from the periodic table) of HCl. I looked it up and its mass is 36.46 g/mol. To get the mols, multiply 15.0 g * 1 mol/36.46g. The grams will cancel out and you will be left with 0.411 mols of HCl.

Since you know that it takes two mols of HCl for every mol of CaCO3, you know that it will take half the number of mols of CaC03, so in this case, about 0.206 mols of CaCO3. To convert that to grams, find the molecular mass, which is 100.087 g/mol. Mulpliying the two cancels out the mols and you are left with 20.6 g of CaC03.

To find the rest of the problem, you basically just do the same thing. Find the amount of mols that are going to be produced and then with the molecular mass, find the weight.

Does that help?

¤CharlDa¤
Feb 6th, 2007, 02:42 AM
I had a friend who had trouble with that. As stupid as it may be, consider the molecules to be sthing you like.

For example:
You want to do 2 bowls of fruit salad. You know that to do two bowls, you need 1 orange for every 2 apples (so if you have 1 apple, you put in 0.5 oranges and do one bowl).

It's the same case in the SO2 + 1/2 O2 --> SO3 reaction. To do one SO3 molecule (fruit salad), you need 1 S02 molecule and half of a 02 molecule.

They tell ya in the problem, that 25g of O2 is useable. When you see grams, you just always make it into moles, because 1 mole of 02 has the same number of molecules than any other grouping. I'm sure you saw how to find the number of moles already (mass/formula weight).

The number you find is the available amount of *oranges* to do your salad. So imagine you have like 10 oranges, and you know that there is 0.5 oranges per bowl. You can conclude that you can do 20 bowls of fruit salad. And to do so, you'll need 20 apples because there is one apple per bowl.

It works the same way in chemistry. If you have, let's say 10 moles of 02, you'll be able to do 20 moles of S03 if you have enough SO2 (20 moles too). The easiest way for me is to find the proportion. Let n = number of moles. so nO2 = number of moles of 02 and nSO3 = number of moles of S03.

First step, you balance your equation. Then just put nO2 = 1/2 (the stoechiometric coefficient) and nSO3 = 1. Now just divide nSO3/n02 = 1/(1/2) = 2. By rearranging the equation, you get: 2nO2 = nS03. When you have that, the moment you find either n02 or nSO3 with the data you have stated in the problem, you are set :D

Hopefully that wasn't too long and clear enough :eek:

Cat's Pajamas
Feb 6th, 2007, 02:57 AM
Ok thank you all, you've guys have been very helpful. but I'm having trouble putting the end pieces together. We are learning to set it up in this box thingy :lol: crossing things out, and this is as far as I seem to be able to get for #1.

15.0g | 1 mol |_________|_________|
------------------------------
_--__| 36.46g | 1 mol |_______|



but also the 36.46 g of HCl, wouldn't you multiply that by 2, since it is 2 moles?

so shouldn't it be:

15.0g | 1 mol |_________|_________|
------------------------------
__--___| 72.92g | 1 mol |_______|


now if someone could just explain the ending part of the question. :wavey:


This information you guys have given me has been :worship:

LeRoy.
Feb 6th, 2007, 03:06 AM
1)
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O

1 MOL + 2 MOL --> 1 MOL + 1 MOL + 1 MOL

FW

CaCO3 = 100
HCl = 36.5
CO2 = 44
CaCl2 = 111

so 100 gms of CaCO3 would react with 2*36.5 gms of HCl to produce 111 gms of CaCl2 and 44 gms of CO2

For 15 gms do the calculations

All you have to do for 1) is solve this

100*15/(2*36.5) gms of CaCO3 would react with 15gms of HCl to produce
111*15/(2*36.5) gms of CaCl2 and 44*15/(2*36.5) gms of CO2

and you have all the answers

Cat's Pajamas
Feb 6th, 2007, 03:19 AM
Ok thanks, now I got it. :D

NyCPsU
Feb 6th, 2007, 03:32 AM
is it bad that im in an organic chemistry class in college and cant even do these problems :help: :o

Cat's Pajamas
Feb 6th, 2007, 03:39 AM
is it bad that im in an organic chemistry class in college and cant even do these problems :help: :o

:haha: Idk ;) But I won't be doing Chemistry in college :bolt:

but I have one more question. Would you use 2 moles instead of 1 mole in the problem since HCl is 2HCl in the balanced equation?

Cat's Pajamas
Feb 6th, 2007, 04:03 AM
by the way, I just realized that for 2. it's actually 100.0g and not 25 ;)

LudwigDvorak
Feb 6th, 2007, 04:52 AM
As far as I know, and if you're in advanced chemistry, only use 1 mole. I'm doing stoichiometry at the moment as well. I understand it a decent bit but it SUCKS taking the class as a senior.

It's generally a sophomore/junior class here, so I'm a genius.

¤CharlDa¤
Feb 6th, 2007, 05:03 AM
Ok, no comment on my fruit bowl allegory :rolleyes: ;)

LudwigDvorak
Feb 6th, 2007, 05:20 AM
Ok, no comment on my fruit bowl allegory :rolleyes: ;)

I tried really hard not to read any equations in here. If it makes you feel better I <3 fruit salad.

SelesFan70
Feb 6th, 2007, 06:11 AM
Unless you're gonna be a chemist...you won't ever use that stuff. ;) :wavey:

Don't sweat it.

mckyle.
Feb 6th, 2007, 10:02 AM
the chem students at my school are doing the same thing right now. :)

hdfb
Feb 6th, 2007, 10:16 AM
this stuff is fun lol.