Stoichiometry Help: NaHC O3 & CaCO3 Data Table

For every 1 mole of calcium carbonate that reacts, you get 1 mole of CO2 produced.2) For every 1 mole of HOAc that reacts, you get 1 mole of CO2 produced.3) Your experimental results gave you (in round numbers, about) 0.0018 moles of CO2 from 0.1 grams of calcium carbonate, and 0.0011 moles of CO2 from 0.1 grams of NaHCO3. You should have gotten 0.0025 moles of CO2 from each of them. You got too little from both, but you got more from the CaCO3 than you did from the Na
  • #1
Mack
9
0
Data Table
.........NaHC O3...CaCO3
1. Mass of micro plate.....2.05g....2.05g
2. Mass of micro plate +salt...3.05g...3.05g
3. Mass of Salt .....1g.....1g
4. Moles of salt.....1/84 mol...1/100 mol
5. Mass of unreacted setup...20.66g....17.9g
6. Mass of reacted setup.....20.58g...17.85g
7. Mass of CO2 lost......08g.....0 5g
8. Moles of CO2 lost.....0018mol...0011 mol
9. Mass of Co2 produced as % of mass of sample 8% 5%

(reacted with vinegar)
Questions:
1. Write balanced equation for each raction.
2. From the stoichiometric relationships in the data table, find the # of moles of CO2 actually produced by each reaction.
3. How did the number of moles of CO2 actually produced by each reaction compare with the number of moles that ideally should be produced?

I don’t get how to do numbers 2 and 3. I believe I did number 2 in 8 on the chart but then what is the ideall number of moles? That’s were I am confused please help
 
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  • #2
use molar mass and use the ratios from the balanced equation
 
  • #3
gradeaswimr said:
use molar mass and use the ratios from the balanced equation

:confused:

well the balanced equation would be for the CaCO3 reaction would be

CaCO3+2HC2H3O2=>Ca(C2H3O2)2+CO2+H2O
molar mass of CaCO3 is about 100g/mol and CO2 is 44g/mol

but i don't no what to do from here or what ratios u r talking about

im sry my teacher didnt explain this at all and I've called all my friends but they don't get it either
 
Last edited:
  • #4
well, Ca has a charge of +2 and the acetate ion has a charge of -1, so you need to rewrite the chemical formula for calcium acetate.

you can ratio the coefficients to link any two pieces in the reaction.

example:

2H2 + O2 -> 2H2O

the ratio ofhydrogen to oxygen is 2:1. the ratio of oxygen to water is 1:2. the ratio of water to hydrogen is 2:1. does that make sense to you?
 
  • #5
the equation isn't balanced
 
  • #6
ya i get that so the ration of CaCO3 and CO2 is like 1:1?

well i think i know what to do but i end up with something like .44mols
and what we did in class for the lab makes out to be .0011mols

i don't get how to get how many grams of CO2 is in 1g CaCO3
if that's what I am supposed to do to get the ideal number of moles?
 
  • #7
gradeaswimr said:
the equation isn't balanced

sry i forgot a 2 i just edited it in so its balanced now sry
 
  • #8
can you show the work with units cause its late and i got my own ap chem final tomorrow.
 
  • #9
read the sections called "molar ratios" and the one called "given grams get grams" - those seem like good tutorials

http://dbhs.wvusd.k12.ca.us/webdocs/Stoichiometry/Stoichiometry.html
 
Last edited by a moderator:
  • #10
i ment i end up with .01 not .44
what i do is i find that CO2 makes up .44g of the 1g ofCaCO3

then i divided .44g by 44g and got .01mols

when i did it in the lab only .o5g evaporated not .44g?

is the work i did correct or the lab correct?
 
  • #11
um...its late, I am tired, those numbers appear so random to me. show all your calculations? just check the tutorials. if you can convert something to moles, you can find basically anything.
 
  • #12
gradeaswimr said:
um...its late, I am tired, those numbers appear so random to me. show all your calculations? just check the tutorials. if you can convert something to moles, you can find basically anything.

molar mass of CO2 is 44g/mol
molar mass of CaCO3 is 100g/mol

1g/100g/mol=x/44g/mol
x=.44g

.44g of CO2 makes up CaCO3?

so .44g divided by 44= .01mols?

idk I am prolly really wrong
 
  • #13
i'm confused. what exactly happened in the lab? did you react calcium carbonate and acetic acid? then wahts "NaHC O3...CaCO3"?
 
  • #14
the grams...grams of what? converting to moles of what?
 
  • #15
we did it with CaCo3 and NaHCO3

it wouldn't let me space the chart out so it would scrunch it all together so i used ... to separate them if i no how to do one i can get the other easily
 
  • #16
so you reacted each one with vinegar?
 
  • #17
yes we reacted each on with vinegar until there was no more bubbles and found how many g of CO2 were evaporated

we used a pipet filled with vinegar and weighed it in with unreacted setup and then weighed it in with reacted setup so that only missing mass was the CO2
 
  • #18
do you know what a limiting reagent is?
 
  • #19
however much acetic acid used was how much reacted. base your stoich on the amount of acetic acid consumed and you should be fine. goodluck & goodnight
 
  • #20
Mack said:
Data Table
.........NaHC O3...CaCO3
1. Mass of micro plate.....2.05g....2.05g
2. Mass of micro plate +salt...3.05g...3.05g
3. Mass of Salt .....1g.....1g
4. Moles of salt.....1/84 mol...1/100 mol
5. Mass of unreacted setup...20.66g....17.9g
6. Mass of reacted setup.....20.58g...17.85g
7. Mass of CO2 lost......08g.....0 5g
8. Moles of CO2 lost.....0018mol...0011 mol
9. Mass of Co2 produced as % of mass of sample 8% 5%

(reacted with vinegar)
Questions:
1. Write balanced equation for each raction.
2. From the stoichiometric relationships in the data table, find the # of moles of CO2 actually produced by each reaction.
3. How did the number of moles of CO2 actually produced by each reaction compare with the number of moles that ideally should be produced?

I don’t get how to do numbers 2 and 3. I believe I did number 2 in 8 on the chart but then what is the ideall number of moles? That’s were I am confused please help

Start by clearly writing each reaction. You have done one already. You can abbreviate acetic acid as 'HOAc' and acetate as 'OAc-'. Calcium actate becomes Ca(OAc)2. Much cleaner, yes?

The second question is not very clearly worded. Read that as "Find the # of moles of CO2 actually produced by each reaction". I don't know how the 'stoichiometric relationships in the data table" can help (or even what it means!). There are no 'stoichiometric relationships' in the data table... only data!

The third question requires those 'stoichiometric relationships' you discovered by answering question #1 (no, those pesky relationships are not in the data table!). Rewriting your answer for the calcium carbonate reaction you have:

CaCO3 + 2 HOAc ----> Ca(OAc)2 + CO2 + H2O

Here is where you will find 'stiochiometric relationships'! You will see that for each mole of CO2 produced you will have started with one mole of CaCO3. You know the weight of the CaCO3 you started with and you can calculate the # of moles. From this you will know how many moles of CO2 you expect to find. Compare that to what you have for mass of CO2 lost (expressed in moles) in your data table.

Now, do the same thing for sodium bicarbonate...
 

1. What is stoichiometry and why is it important?

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is important because it allows us to predict the amount of products that will be formed from a given amount of reactants, and vice versa. This helps in understanding and controlling chemical reactions in industries and laboratories.

2. What is the purpose of the NaHC O3 & CaCO3 data table?

The data table is used to record the amounts of reactants (NaHC O3 and CaCO3) and products (Na2CO3 and H2O) in a chemical reaction. This data is then used to calculate the stoichiometric ratios and determine the limiting reactant in the reaction.

3. How do you determine the limiting reactant using the data table?

To determine the limiting reactant, you need to compare the amount of each reactant used in the reaction with the stoichiometric ratios. The reactant that has the smallest ratio is the limiting reactant, as it will be completely consumed in the reaction.

4. Can you explain how to calculate the theoretical yield using stoichiometry?

To calculate the theoretical yield, you need to first determine the limiting reactant. Then, using the stoichiometric ratios, you can calculate the moles of the product that would be formed from the limiting reactant. Finally, using the molar mass of the product, you can convert the moles to grams to get the theoretical yield.

5. How does stoichiometry help in chemical reactions?

Stoichiometry helps in chemical reactions by allowing us to predict and control the amount of products that will be formed from a given amount of reactants. It also helps in optimizing the use of reactants, reducing waste, and ensuring the efficiency of the reaction. Additionally, stoichiometry can also be used to determine the purity of a product and identify any impurities in a reaction.

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