Bianca chlorine is the limiting reactant because you can see that in the drawing that you did.In the picture you will see there is two hydrogen for every chlorine. the excess reactant would hydrogen because there is more of it.
only 2H remains because for every chlorine there is two hydrogen. if you look at the picture that is given you see that there is more hydrogen then chlorine so therefor the excess is hydrogen. if you draw it out you will that there is two hydrogen left
For number two, part a. use dimensional analysis (100g Na)(1 mol Na/22.98977 g Na) = 4.35 mol Na I rounded my answer but it's originally 4.3479760785.
(100 g Fe2O3) (1 mol Fe2O3/ 159.6922 g Fe2O3)=.626 mol Fe2O3
Then, divide (4.35 mol Na/.626 mol Fe2O3)= 6.95 mol Na/1 mol Fe2O3 Compare this to the molar ratio from the equation which is (6 mol Na/1 mol Fe2O3) The excess amount is Na (Sodium) because we have about 6.95 of it when we only need 6 mol of it. The limited amount is Fe2O3 (Iron Oxide).
I hope this is right and helps. If I made a mistake or a typo, I apologize.
To find the limiting and excess reactant, first you have to balance the equation, which you already did. Then you have to convert the reactants to moles. for Na, 100g of Na is used, so it'll be 100g Na (1 mol Na/ 22.98977g Na) = 4.3498 mol Na. (You get the mass from the reference table.) For Fe2O3, 100g of Fe2O3 is used, so it'll be 100g Fe203 (1 mol Fe2O3/159.6922 Fe2O3) = .626 mol Fe2O3. Now you can compare the molar ratios. 4.3498 mol Na/.626 mol Fe2O3 = 6.9486 mol Na/1 mol Fe2O3. Only 6 moles of Na is needed for this reaction, and you have 6.9486, so Na is the excess reactant,and Fe2O3 is the limiting reactant.
LJ, you could use the method that Grossman expained to us. For each compoud use a letter, CO2-a H2o-b C6H12O6-C O2-d Then, from there, go by each elemnt C- 1a=6c there is 1 carbon in the first compoud, then u put = becuase c and d are the products. You do that for all the elements O- 2a+1b=6c+2d H-2b=12c
Then you replace a with 6 we usually replace it with a 2, but in this case, it wont work becuase the coafficients are bound to be larger numbers then 2, because Glucose (c6,h12,02) is a big compound.
ok so if a is 6, then c has to be 1 for the to equal if c is 1 then b is 6,
from there on u just plug in the numbers and figure out a number for each a,b,c,d
what ever the numbers for a,b,c, d are then those are the coefficients for each compoud.
NickO u can delete a comment when u type in something and publish it there is a little trash can next to the date and time click it and it will ask u if u wanna remove it.
I dont get 1b. of the hw How do u knoe what the Limting reactant (reagent) is 4 the problem?
ReplyDeleteBianca
ReplyDeletechlorine is the limiting reactant because you can see that in the drawing that you did.In the picture you will see there is two hydrogen for every chlorine. the excess reactant would hydrogen because there is more of it.
ok but now i also dont get what the excess reactant is nd how much of it remains
ReplyDeleteBianca
ReplyDeleteonly 2H remains because for every chlorine there is two hydrogen. if you look at the picture that is given you see that there is more hydrogen then chlorine so therefor the excess is hydrogen. if you draw it out you will that there is two hydrogen left
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ReplyDeleteNick, please hold the political commentary
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ReplyDeleteThis comment has been removed by a blog administrator.
ReplyDeletehow do you find the limiting & excess reactants for #2 after you balanceit
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ReplyDeleteI don't understand number 2, is it suppose to be 100g of Na2O3 or Fe2O3?
ReplyDeleteI sent an email correcting that. Yes it is supposed to be Fe2O3
ReplyDeleteNick O.
ReplyDeletePlease try to keep those random outbursts to a minimum and use the blog for help. Thanks.
(Just like Mr Grossman sais ) :]
Ok,
for 1a in the second box you have to draw 1 Clorine atom paired up with two Hydrogen atoms
How do I know that?
From the equation :
H2+Cl2 -->2HCl
What u drew is wrong !
Hopefully i helped :)
+1 for Marina hehe
LJ and Jamie,
ReplyDeleteFor number two, part a. use dimensional analysis
(100g Na)(1 mol Na/22.98977 g Na) = 4.35 mol Na
I rounded my answer but it's originally 4.3479760785.
(100 g Fe2O3) (1 mol Fe2O3/ 159.6922 g Fe2O3)=.626 mol Fe2O3
Then, divide (4.35 mol Na/.626 mol Fe2O3)= 6.95 mol Na/1 mol Fe2O3
Compare this to the molar ratio from the equation which is (6 mol Na/1 mol Fe2O3)
The excess amount is Na (Sodium) because we have about 6.95 of it when we only need 6 mol of it.
The limited amount is Fe2O3 (Iron Oxide).
I hope this is right and helps. If I made a mistake or a typo, I apologize.
This comment has been removed by a blog administrator.
ReplyDeleteLJ,
ReplyDeleteTo find the limiting and excess reactant, first you have to balance the equation, which you already did. Then you have to convert the reactants to moles. for Na, 100g of Na is used, so it'll be 100g Na (1 mol Na/ 22.98977g Na) = 4.3498 mol Na. (You get the mass from the reference table.) For Fe2O3, 100g of Fe2O3 is used, so it'll be 100g Fe203 (1 mol Fe2O3/159.6922 Fe2O3) = .626 mol Fe2O3. Now you can compare the molar ratios. 4.3498 mol Na/.626 mol Fe2O3 = 6.9486 mol Na/1 mol Fe2O3. Only 6 moles of Na is needed for this reaction, and you have 6.9486, so Na is the excess reactant,and Fe2O3 is the limiting reactant.
aah Awista!
ReplyDeleteI'm sorryyy Jamie!
ReplyDeletelol can you help me with 2b? :)
ReplyDeleteThis comment has been removed by the author.
ReplyDeleteHAHAHA xD
ReplyDeleteits "dashuri "
your welcome :] Do u have any other questions?
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ReplyDeleteThis comment has been removed by the author.
ReplyDeletehow do you balance this equation CO2+H2O--- C6H12O6+O2
ReplyDelete6C02+6H20->C6H12O6+6O2
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ReplyDeleteLJ, you could use the method that Grossman expained to us. For each compoud use a letter,
ReplyDeleteCO2-a
H2o-b
C6H12O6-C
O2-d
Then, from there, go by each elemnt
C- 1a=6c
there is 1 carbon in the first compoud, then u put = becuase c and d are the products. You do that for all the elements
O- 2a+1b=6c+2d
H-2b=12c
Then you replace a with 6
we usually replace it with a 2, but in this case, it wont work becuase the coafficients are bound to be larger numbers then 2, because Glucose (c6,h12,02) is a big compound.
ok so if a is 6, then c has to be 1 for the to equal
if c is 1 then b is 6,
from there on u just plug in the numbers and figure out a number for each a,b,c,d
what ever the numbers for a,b,c, d are then those are the coefficients for each compoud.
the final answer would be
6CO2+6H2O-->C6H12O6+602
Who gets 3c, the reactants being limited to what?
ReplyDeletei dont get the question
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ReplyDeleteNickO
ReplyDeleteu can delete a comment when u type in something and publish it there is a little trash can next to the date and time click it and it will ask u if u wanna remove it.
This comment has been removed by a blog administrator.
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ReplyDeletecan someone help me with 2b i dont understand what i have to do
ReplyDeleteDaniel ,
ReplyDeleteWhat I got for 2b is the following :
Fe= 36*2 = 72 (mass of Fe)
I got the 2 from the equation
The coefficient of Fe is 2
so i multiplied the mass of Fe(s) by 2
aii thanks marina
ReplyDeleteno problem :]
ReplyDelete