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Mole Concept ques!

D

Dean713

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soumayya

soumayya

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s07 qp3 #7 (d)
100 g of fat reacts with 86.2 g of iodine.
884 g of fat reacts with [(86.2/100)*884] = 762.008 g of iodine.
One mole of fat reacts with (762.008/254) = 3 moles of iodine molecules.
Number of double bonds in one molecule of fat is 3.
 
soumayya

soumayya

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s09 qp 3 # 9(a)
------------------Mg : N
mass------------ 72 : 28
Divide by Ar----72/24 : 28/14
No. of moles---3 : 2

Empirical formula= Mg3N2
 
D

Dean713

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soumayya said:
s09 qp 3 # 9(a)
------------------Mg : N
mass------------ 72 : 28
Divide by Ar----72/24 : 28/14
No. of moles---3 : 2

Empirical formula= Mg3N2
Click to expand...

isnt Mg3N2 the moles but u need to find mole ratio
so dt u hav to divide each by the smallest mole
so shouldnt it be Mg2N???
 
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Dean713

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soumayya said:
s07 qp3 #7 (d)
100 g of fat reacts with 86.2 g of iodine.
884 g of fat reacts with [(86.2/100)*884] = 762.008 g of iodine.
One mole of fat reacts with (762.008/254) = 3 moles of iodine molecules.
Number of double bonds in one molecule of fat is 3.[/quot
Click to expand...
 
soumayya

soumayya

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Dean713 said:
soumayya said:
s09 qp 3 # 9(a)
------------------Mg : N
mass------------ 72 : 28
Divide by Ar----72/24 : 28/14
No. of moles---3 : 2

Empirical formula= Mg3N2
Click to expand...

isnt Mg3N2 the moles but u need to find mole ratio
so dt u hav to divide each by the smallest mole
so shouldnt it be Mg2N???
Click to expand...
yh...but in this ques u already get the mole ratio in whole numbers..
still..suppose u divide by the smallest mole, u get 1.5 for Mg and 1 for N..
1.5 : 1 = 3:2
u can't just round the figures unless it is very close to a whole number...
 
SuperXDE

SuperXDE

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In other words , write down the equation , balance it , and write the balanced stuff in moles , just like this

Respiration : 1 NaOH + 1 HCl ---> 1 NaCl + 1 H2O , 1 Sodium Hydroxide Mole + 1 Hydrochloric Acid Mole ---- > 1 Sodium Chloride mole + 1 Water mole , if it is something like 2Mg + O2 ---> 2 MgO , then it is like 2 Magnesium Mole + 1 oxygen Mole ---- > 2 Magnesium Oxide moles , it is more like Algebra in that part , like a ratio , how much moles of Magnesium required to produce a single mole of Magnesium oxide. I remember I did write something about it in the Stickies , if you still don't get it , Tell us.
 
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Dean713

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soumayya said:
Dean713 said:
soumayya said:
s09 qp 3 # 9(a)
------------------Mg : N
mass------------ 72 : 28
Divide by Ar----72/24 : 28/14
No. of moles---3 : 2

Empirical formula= Mg3N2
Click to expand...

isnt Mg3N2 the moles but u need to find mole ratio
so dt u hav to divide each by the smallest mole
so shouldnt it be Mg2N???
Click to expand...
yh...but in this ques u already get the mole ratio in whole numbers..
still..suppose u divide by the smallest mole, u get 1.5 for Mg and 1 for N..
1.5 : 1 = 3:2
u can't just round the figures unless it is very close to a whole number...
Click to expand...

But still they're askin 4 mole ratio not moles..bu i guess tht makes sense..
 
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Dean713

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SuperXDE said:
In other words , write down the equation , balance it , and write the balanced stuff in moles , just like this

Respiration : 1 NaOH + 1 HCl ---> 1 NaCl + 1 H2O , 1 Sodium Hydroxide Mole + 1 Hydrochloric Acid Mole ---- > 1 Sodium Chloride mole + 1 Water mole , if it is something like 2Mg + O2 ---> 2 MgO , then it is like 2 Magnesium Mole + 1 oxygen Mole ---- > 2 Magnesium Oxide moles , it is more like Algebra in that part , like a ratio , how much moles of Magnesium required to produce a single mole of Magnesium oxide. I remember I did write something about it in the Stickies , if you still don't get it , Tell us.
Click to expand...

wats d "stickies"?
anyways i get it from wat soumayya said...
so moles n mole ratio is like the same thing if u
already have whole numbers like in tht case??...
 
SuperXDE

SuperXDE

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Stickies on a forum are topics which are thought to be of importance by the moderators , and they were put in the first of the list , check the IGCSE Forum , you'll see that there are three topics in the head , Click here to go to the topic I was talking about.

One thing else ... you give up quickly on the mole question , just write the question and write your Exam-style answer , and seek correction of your answer , thats better than just asking how. :)
 
Sara.H

Sara.H

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soumayya said:
s07 qp3 #7 (d)
100 g of fat reacts with 86.2 g of iodine.
884 g of fat reacts with [(86.2/100)*884] = 762.008 g of iodine.
One mole of fat reacts with (762.008/254) = 3 moles of iodine molecules.
Number of double bonds in one molecule of fat is 3.
Click to expand...
It Helps a lot ..thank you dear :D
 
Sara.H

Sara.H

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Dean713 said:
isnt Mg3N2 the moles but u need to find mole ratio
so dt u hav to divide each by the smallest mole
so shouldnt it be Mg2N???
Click to expand...
Was just having the same confusions .-. :p
Soumyya u r a true helper xD now finally I've get it :))
 
Sara.H

Sara.H

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Dean713 said:
plz help!

s07 qp3 #7 (d)
http://www.xtremepapers.net/CIE/Cambrid ... 7_qp_3.pdf
Ques with one mole of fat reacts with....

s09 qp 3 # 9(a)
http://www.xtremepapers.net/CIE/Cambrid ... 9_qp_3.pdf
(1st varient)
the answer im gettin is Mg2N but i dnt understand how the markscheme has Mg3N2??

s06 qp3 #7(d)
http://www.xtremepapers.net/CIE/Cambrid ... 6_qp_3.pdf
how do u calculate maximum moles??

how do u calculate to find the excess of a substance???
Click to expand...
s06 qp3 #7(d)
Maximum moles re always simply calculated by the ratio
So propene : 2 idopropane
1:1
0.0333 0.0333. It would absolutely have the same no.of moles
 
Sara.H

Sara.H

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Dean713 said:
plz help!

s07 qp3 #7 (d)
http://www.xtremepapers.net/CIE/Cambrid ... 7_qp_3.pdf
Ques with one mole of fat reacts with....

s09 qp 3 # 9(a)
http://www.xtremepapers.net/CIE/Cambrid ... 9_qp_3.pdf
(1st varient)
the answer im gettin is Mg2N but i dnt understand how the markscheme has Mg3N2??

s06 qp3 #7(d)
http://www.xtremepapers.net/CIE/Cambrid ... 6_qp_3.pdf
how do u calculate maximum moles??

how do u calculate to find the excess of a substance???
Click to expand...

U should firstly now how to indicate the limiting reactent
By the following 3 steps: 1-moles of first reactant 2-moles of 2nd reactant 3-Ratio
4 ex: 2p (20 g) + 3cl2 (50 g) > 2pcl3
Calculate the moles fisrt which 's mass/Mr
You'll get p =0.65 moles Cl2 =0.70 moles
Now Ratio : 2 mol>3mol
0.65
0.65*3/2 ~ (0.975) which's the moles needed
But we have 0.70 moles which is less than needed for complete reaction therefore cl2 is the limiting reactant

Or

2 mol>3mol
?> 0.70
0.47 mol needed to complete reaction ..but we have 0.65 mol ,which is more than needed so p is in excess

hope it's helps^_^
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