1-Convert mass of reactants into moles 2-Compare their current molar ratio to the molar ratio in the balanced equation Which ever reactant is smaller is the limiting. 3-Use moles of limiting reactant to see how much product is formed in grams.
If you need further explanation then feel free to ask =]
1a) You start with 24.8 g of CaCO3 Dimensional Analysis: -1 mol CaCO3/(40.08 + 12.0111+ 15.9994 x 3)g CaCO3 - 1 mol CaO/ 1 mol CaCO3 -(40.08 + 15.9994)g CaO/ 1 mol CaO My answer is 13.89528 grams of CaO. Rounded to 14. Correct me if I'm wrong.
Ehhh You mean 1b ? LJ, For #2 You take 1.87 grams of Al and convert it to moles. Then you do the same with Copper (II) Sulfate, meaning convert 9.65 to moles. Use your solutions to find the limiting reactant. (To do it: set the numbers of moles as a fraction and set it equal to the ratio shown by the chemical equation after the proportions have the common denominator.) Use grams of the limiting reactant which is already given to find the mass of copper. Put the mass of copper under the actual mass of copper which is 3.65. And that's it!
When finding the limiting and excess reactants, take the amount which is given to you for example I'll use what is given in number three, if they give you 100 g of a certain reactant and 100 g for another reactant. Convert each reactant which is now in grams to moles. (100 g SO2)(1 mol SO2/64 g SO2) The number of grams is found in the reference table. (100g NaOH)(1 mol NaOH/39.997 g NaOH) Then divide the number of moles for each reactant. (1.56 mol SO2/2.5 mol NaOH)= (.624 mol SO2/ 1 mol NaOH) Now, find the molar ratio of these two reactants from the equation. (1 mol SO2/ 2 mol NaOH)which you get from the coefficents in front of the reactants in the equation. This can get simplified to (.5 mol SO2/ 1 mol NaOH) To determine which one is the limiting reactant, look at what you have which is .624 mol SO2 and you need only .5 mol SO2 so the excess amount is SO2 because you have more than is necessary and the limiting reactant is NaOH.
the question will tell you the actual yield. in order to find the theoretical yield, you must find the limit reactant first(exa: question #2. after calculation, Al is the limited.). then use the limited reactant to find the theo yield.
find the theo yield by(use #2 as ex) take the limited reactant ammount(1.87g Al) turn it into mole(1.87g Al*(1mole Al/26.98154(molar mass))).Then times the result with the molar radio between this limited reactant and the product you need to find for the theo yield(Cu)(times by 3mole Cu/2mole Al).After this step, you will get the mole of the product you need to find for the theo yield(Cu). Change it from mole to mass by times it by the molar mass (molar mass of Cu). LASTLY THIS IS YOUR THEO YIELD FOR Cu.
i am lost someone please help me with #1
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ReplyDeleteThis comment has been removed by the author.
ReplyDeletehow do you find the percentyield of it
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ReplyDeleteNick O.
ReplyDeleteThe steps that you need to take to solve # 3 :
1-Convert mass of reactants into moles
2-Compare their current molar ratio to the molar ratio in the balanced equation
Which ever reactant is smaller is the limiting.
3-Use moles of limiting reactant to see how much product is formed in grams.
If you need further explanation then feel free to ask =]
-Marina
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ReplyDeleteLJ
ReplyDeleteI'll try my best to help you with #1 A and B
you need to calculate the theoritical yield
so u start with what is given to you the (24.8grams of CaC)
( 1mol CaC/52g CaC ) <--the mass of CaC
(1mol CaO/1mol CaC)
(55gCaC/1molCaC)<--mass of CaO
you solve and the answer is 26.2
In order to find the PERCENT YIELD :
ReplyDeleteyou use the formula (actual yield/theoritical yield) * 100
you start with the 13.1grams -->the actual yield
&& you divide it by the theoritical yield 26.2
*100
that comes out to be 50%
i hope this is right =P (at least it looks right)
Marina motra, it starts of with 24.8g of CaCO3(Calcium Carbonate), not CaC
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ReplyDeleteyess my motra is right ! :]
ReplyDeletegot confused x[
so wait so then is it 14.06 ?
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ReplyDeletecan someone help me with #2&3 please i neeed help!!!!!!!!
ReplyDelete1a)
ReplyDeleteYou start with 24.8 g of CaCO3
Dimensional Analysis:
-1 mol CaCO3/(40.08 + 12.0111+ 15.9994 x 3)g CaCO3
- 1 mol CaO/ 1 mol CaCO3
-(40.08 + 15.9994)g CaO/ 1 mol CaO
My answer is 13.89528 grams of CaO.
Rounded to 14. Correct me if I'm wrong.
14 is right!! that's what i got too
ReplyDelete2b is
13.1 /14 = 0.9357 * 100 =
93.57..
u round it up and it should be 94 %
Ehhh You mean 1b ?
ReplyDeleteLJ,
For #2 You take 1.87 grams of Al and convert it to moles. Then you do the same with Copper (II) Sulfate, meaning convert 9.65 to moles. Use your solutions to find the limiting reactant. (To do it: set the numbers of moles as a fraction and set it equal to the ratio shown by the chemical equation after the proportions have the common denominator.)
Use grams of the limiting reactant which is already given to find the mass of copper. Put the mass of copper under the actual mass of copper which is 3.65. And that's it!
Edit: Divide and times one hundred and then add the percentage sign.
ReplyDeletei am still confused on how to find the limiting & excess reactants
ReplyDeleteplease help!!!!!
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ReplyDeletei dont get it when you put it in a ratio form and hoe to get itt from that
ReplyDeletein nnumber 2 what are u using the 1.87 in your calculation
ReplyDeleteLJ,
ReplyDeleteWhen finding the limiting and excess reactants, take the amount which is given to you for example I'll use what is given in number three, if they give you 100 g of a certain reactant and 100 g for another reactant. Convert each reactant which is now in grams to moles. (100 g SO2)(1 mol SO2/64 g SO2) The number of grams is found in the reference table. (100g NaOH)(1 mol NaOH/39.997 g NaOH) Then divide the number of moles for each reactant. (1.56 mol SO2/2.5 mol NaOH)= (.624 mol SO2/ 1 mol NaOH) Now, find the molar ratio of these two reactants from the equation. (1 mol SO2/ 2 mol NaOH)which you get from the coefficents in front of the reactants in the equation. This can get simplified to (.5 mol SO2/ 1 mol NaOH) To determine which one is the limiting reactant, look at what you have which is .624 mol SO2 and you need only .5 mol SO2 so the excess amount is SO2 because you have more than is necessary and the limiting reactant is NaOH.
how can you tell if it is a actual or theo yield? how can you find it if it not given in the question? please heeeelp!!!!!!
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ReplyDeleteLJ, post your comments in the take home exam section
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ReplyDeletethe question will tell you the actual yield. in order to find the theoretical yield, you must find the limit reactant first(exa: question #2. after calculation, Al is the limited.). then use the limited reactant to find the theo yield.
ReplyDeleteLJ,
ReplyDeletefind the theo yield by(use #2 as ex) take the limited reactant ammount(1.87g Al) turn it into mole(1.87g Al*(1mole Al/26.98154(molar mass))).Then times the result with the molar radio between this limited reactant and the product you need to find for the theo yield(Cu)(times by 3mole Cu/2mole Al).After this step, you will get the mole of the product you need to find for the theo yield(Cu). Change it from mole to mass by times it by the molar mass (molar mass of Cu). LASTLY THIS IS YOUR THEO YIELD FOR Cu.