Grade 11 Stoichiometry help
So I'm stuck on this question (I got the answer for the first part of the question, but not the second) and I was wondering if anyone knew how to complete the second part?
The chemical formula:
2KCl(aq) + 1Pb(NO3)2(aq) ----> 1PbCl2(s) + 2KNO3(aq) + 215 kJ
The question:
A 59.0% pure sample of Pb(NO3)2 which has a mass of 6.5g reacts with excess KCl. What mass of KNO3 could I produce? If I run this experiment in a 2.0L vessel, what will be the molar concentration of the resulting KNO3(aq) solution?
May 23, 2013
8 Comments • Newest first
@Eruditez: This is very simple stoich though o-o if you plan on taking the AP test, i suggest you start studying. not knowing how to do simple calculations such as these will not even fetch you a 2 on the AP test...
@LiliKoby: oh I didn't see what you put, and this is the only question that I didn't seem to get, I did the rest of my work, it only this question.
@Eruditez: Correct my calculations if anything is wrong lol, but it should be correct unless i misread the question.
Oh god, I did this in grade 10
It was torture
I usually had @NonSonoFronz to help me
Oh and @oyster helped me too
Alright, thanks guys
@silverphoenix
@lordwired
If you guys aren't busy, can you try this question and tell me what answer you get for both parts? You don't need to show any work or anything.
sample has 6.5g. 59% of 6.5=3,835 grams Pb(NO3)2. 3.835/331=.0115 moles pb(no3)2.(331g is molar mass of lead nitrate) there is 2 moles of KNO3 for every 1 mole lead nitrate reacted, so there is a total of .023 moles of potassium nitrate, or 2.34 grams. (molar mass of potassium nitrate is 101g/mol) concentration is moles per liter, so .023/2 results in a .0115 mol/L concentration.
OT: I helped, but you should really do your own homework. and this is by no means 11 grade chem. i could easily do these kinds of questions freshman year. you should really just figure it out.
Molar concentration = Moles / Volume (in Liters)
If you've figured out the mass of KNO3, divide it by the molar mass of KNO3. Following that, divide it by 2 L which gives you the appropriate mol/L units