Calculating Mass of Galena Converted to Lead Oxide: Quick Heat Question

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The discussion focuses on calculating the mass of galena (PbS) converted to lead (II) oxide (PbO) when 975 KJ of heat is liberated during the reaction. The initial approach suggests that 975 KJ is not per mole, leading to a calculation of 1.18 moles of PbS, which corresponds to 282 grams. Participants express concerns about the interpretation of heat units and whether the specific heat value of 827.4 KJ is correct. There is confusion regarding the correct application of formulas, with some participants questioning the necessity of using the provided reaction equation. Ultimately, the conversation highlights the importance of clarity in unit conversions and the correct application of thermodynamic principles.
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Problem: Galena (PbS) is heated in air to form lead (II) Oxide.

2PbS(s) + 3O2 (g) --- 2PbO (s) + 2SO2 (g)

What mass of galena is converted to lead oxide if 975 KJ of heat are liberated?

My Approach:

975 KJ x (1mol/827.4KJ) = 1.18 mol

1.18 mol PbS x (239.3g PbS/1mol PbS) = 282g PbS

My concern is whether the 975KJ is in fact 975 KJ/mol...if it is, then it's going to change the answer quite a bit! Thanks
 
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Well if in the problem you are given 975KJ i would assume that its just KJ not KJ/mol.
 
Stevedye56 said:
Well if in the problem you are given 975KJ i would assume that its just KJ not KJ/mol.

so you think my approach is correct?
 
Is the the specific heat 827.4KJ?
 
^^^yes it is...damn i forgot to put that on here oops
 
cheechnchong said:
^^^yes it is...damn i forgot to put that on here oops

Ok. It looks like the correct procedure assuming your multiplication is right. At first i thought it was q=smdeltaT but there was no temperature so i guess it couldn't be that.
 
Stevedye56 said:
Ok. It looks like the correct procedure assuming your multiplication is right. At first i thought it was q=smdeltaT but there was no temperature so i guess it couldn't be that.

hmmm, my TA marked it wrong...right now I am guessing whether we have to utilize the equation they provide? what do you think?
 
I think i found one of the mistakes. KJ/mol is not mol/KJ

975 KJ x (1mol/827.4KJ) = 1.18 mol
 
Stevedye56 said:
I think i found one of the mistakes. KJ/mol is not mol/KJ

975 KJ x (1mol/827.4KJ) = 1.18 mol

ummm i don't think it's wrong really...i actually think this is right? i think it's right because when you figure out the mol amount (this is an example) from .3g O2, you usually divide it by 32 g/mol right? that's what i thought when i approached this problem...let me know if I am wrong
 
  • #10
Hi,
I need help in answering this question...
Why would the scientific community resist the concept of a nucleus full of positive charges and what observation helped them explain adn accept this design?
Thanks,
v
 
  • #11
vfdismer001 said:
Hi,
I need help in answering this question...
Why would the scientific community resist the concept of a nucleus full of positive charges and what observation helped them explain adn accept this design?
Thanks,
v

Why are you posting this in another thread?
 
  • #12
vfdismer001 said:
Hi,
I need help in answering this question...
Why would the scientific community resist the concept of a nucleus full of positive charges and what observation helped them explain adn accept this design?
Thanks,
v

well i can only think of ways where the concept is applicable. The scientific community who resist the positive charge of the nucleus obviously haven't performed rutherford's gold foil experiment. The only way i can see them disagree is because of magnetic fields disregarding the nucleus--the layer of a magnet is what causes the (+) and (-), vice versa--probably their belief (but i think the nucleus has something to do with the magenetic reaction). I dunno, this answer is straight off guessing...
 
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