# Can someone please explain the calculation questions? (1 Viewer)

#### emmabravica

##### New Member
The first one was the 5 marker where 12.?? g of lead was reacted with 0.1L 1M silver nitrate and then you had to find the final amount of moles of Pb(s), Pb2+, Ag(s), Ag+, NO3-. I got the balanced equation but wow did not know how to do the moles so just guessed! Probably a simple basic question but overthinked it

Same with the 4 marker how 0.249 g of ethanol (or some amount) was combusted to raise the temperature of an oily liquid to 120 degrees (or something of the sort). Thanks! Any ideas?

#### drunkenmonkee37

##### Member
The first one was the 5 marker where 12.?? g of lead was reacted with 0.1L 1M silver nitrate and then you had to find the final amount of moles of Pb(s), Pb2+, Ag(s), Ag+, NO3-. I got the balanced equation but wow did not know how to do the moles so just guessed! Probably a simple basic question but overthinked it

Same with the 4 marker how 0.249 g of ethanol (or some amount) was combusted to raise the temperature of an oily liquid to 120 degrees (or something of the sort). Thanks! Any ideas?
This...
I overthought te 5 marker one but then realised it was a basic procedure to calculate the moles
I have mixed feelings about the nekk one

#### superSAIyan2

##### Member
First write out the equation: Pb + 2AgNO3 ----> Pb(NO3)2 + 2Ag
Initial moles of Pb = (20.72g) / (207.2g/mol) = 0.1moles
Initial moles of Ag+ = 0.1 L x 1 mol/L = 0.1moles
now for a complete reaction n(Pb):n(Ag+) = 1:2
therefore all 0.1 moles of Ag+ has reacted to form Ag solid and only 0.05 moles of Pb has reacted to form Pb(2+)
Therefore final concentrations are : Ag+ = 0, Ag = 0.1, Pb = 0.05, Pb(2+) = 0.05, NO3(-) = 0.1 (since nitrate is spectator)

#### ocatal

##### Active Member
For the oily liquid question, what was the final temperature? 40 or 50?

#### Talia95

##### New Member
I was stuck with it too, just ended up trying to work out the molar ratios. Ended up with the Pb(s) and Ag+ being 0, so I think I made a mistake somewhere...
With the ethanol one, read off the graph for the temperature change of the oily liquid. Found the moles of ethanol used and multiplied it by the heat of combustion of ethanol to find the heat released. Then just used algebra to find the heat capacity, by using the -mCAT formula

#### drunkenmonkee37

##### Member
General consensus says 40 at my school, so i guess i prolly fucked up big time

#### superSAIyan2

##### Member
for the next question
find moles of ethanol n(ethanol) = 0.249/46... = 0.005...moles (i cant remember the exact numbers)
Find heat released by this sample q= 0.005 x 1367 = 7... Kj

now mass oil = 0.12kg and temperature change is 30degrees (it is not 20 because final temp occurs when heating stops)
use q=mCT where m=0.12, C=specific heat of oil and T=30
you should get C=2.13 Kl/kg/degrees or C=2.13 J/g/degrees

#### hamstar

##### Member
For the oily liquid question, what was the final temperature? 40 or 50?
50, thats when the ethanol stopped burning.

#### DiValdez

##### Member
for the heat capacity i got 10^-3 x 2

#### superSAIyan2

##### Member
^ did you convert your mass to kilograms OR your heat to joules

specific heat capacity is usually expressed as X Kj/Kg.degrees or X J/g/degrees

EDIT: just saw the back of the periodic table. the heat capacity can also be written as J/kg/ degrees

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#### hamstar

##### Member
it was a limiting reagent question. u are left with 0.05 moles of Pb, 0 Ag+, 0.05 moles of Pb2+, 0.1 moles Ag. I think this was the question not really sure.

#### Frie

##### Wannabe Inventor
I got 2.13, wasnt sure about units, so I just copied the one on back of periodic table for water, but instead of 4.18 i replaced with 2.13

So 2.13 x10^3 J Kg^-1 K^-1

#### hamstar

##### Member
I got 2.13, wasnt sure about units, so I just copied the one on back of periodic table for water, but instead of 4.18 i replaced with 2.13

So 2.13 x10^3 J Kg^-1 K^-1
+1

#### babberz

##### New Member
First write out the equation: Pb + 2AgNO3 ----> Pb(NO3)2 + 2Ag
Initial moles of Pb = (20.72g) / (207.2g/mol) = 0.1moles
Initial moles of Ag+ = 0.1 L x 1 mol/L = 0.1moles
now for a complete reaction n(Pb):n(Ag+) = 1:2
therefore all 0.1 moles of Ag+ has reacted to form Ag solid and only 0.05 moles of Pb has reacted to form Pb(2+)
Therefore final concentrations are : Ag+ = 0, Ag = 0.1, Pb = 0.05, Pb(2+) = 0.05, NO3(-) = 0.1 (since nitrate is spectator)
i got this too, hope its right

#### kimjuliana

##### Member
haha I made a really epic silly. For some stupid reason I wrote Ag+ = 0, Ag = 0.1, Pb = 0.5, Pb(2+) = 0.5, NO3(-) = 0.1
instead of Ag+ = 0, Ag = 0.1, Pb = 0.05, Pb(2+) = 0.05, NO3(-) = 0.1
How many marks would i lose?? :'(

#### superSAIyan2

##### Member
^ in your working out if you wrote 0.05 moles i think you might only lose 1 mark.

#### kimjuliana

##### Member
i didn't write 0.05 moles... arghhh! and btw what was the question worth?

#### ocatal

##### Active Member
i didn't write 0.05 moles... arghhh! and btw what was the question worth?
5.

Isn't Pb 0?

#### superSAIyan2

##### Member
^ no. Lead is in excess in the reaction, isnt it?