Hard chemistry problem- can't figure it out its killing me- show work please?
Answer:
weight percentage of KCl = 57.604%
weight percentage of NaCl = 42.396% (be sure to read the "Added" section for why this differs from 5) below)
1) Use the amount of AgCl to find the amount of Cl. You can do this by taking their weights (per mole) to find the ration of one mole of Cl to one mole of AgCl and then multiplying that ratio by the weight of AgCl. According to the Los Alamos website's periodic table, the mole weights of the four elements are: Ag = 107.9g, Cl = 35.45g, K = 39.10g and Na = 22.99g. So: 35.45 / (107.9 + 35.45) * 2.1476g = 0.531095g of Cl.
2) Now you can find the amount of K + Na in the 1g mixture of NaCl and KCl by subtracting the amount of Cl from the 1g: 1g - 0.531095g = 0.468905g of K and Na.
3) Next, find the ratio of the K and Na in the 0.468905g and multiply by the 0.468905g to find the weight of one, then subtract from 0.468905g to find the weight of the other: 39.1 / (39.1 + 22.99) * 0.468905g = 0.295284g of K and therefore 0.468905g - 0.295284g = 0.173621g of Na.
4) Now combine the weight of each with its share of the weight of Cl. This is fairly simple as the molecular formulae are both 1:1 ratios and you can assume the Cl does not combine with either preferentially. So: 0.295284g + (0.531095 / 2)g = 0.5608315g of KCl and 1g - 0.5608315g = 0.439175g of NaCl.
5) Now finding the weight percentages of each is fairly simple since you will be dividing by 1: 0.5608315g / 1g = 56.083% KCl and 1 - 56.083% = 43.917% NaCl.
6) Use five significant digits even thought the weight of the original sample was given to one digit. There's little point to giving their weight percentages to one digit: 100% KCl and 0% NaCl seems unlikely to thrill a teacher... Since the other specified value is to five digits (2.1476g of AgCl), it seems suitable to use it for the significant digits part of the answer.
Added:
Did some fiddling with a spreadsheet to find the maximum KCl weight percentage. It works out to slightly more than half and half uptake of the Cl. At 0.30212g of K, the KCl weight percentage is 57.604%. "Optimizing" for NaCl in the same manner, we find it comes out the same. Of course, that figures, doesn't it? (When I say optimize, I mean hitting the 1g mark pretty exactly which it turns out that assuming 50-50 uptake does not (it comes out to slightly over the 1g; I had assumed that was rounding effect, but it seems not by these results.) The 50-50 uptake seemed a reasonable simplification from the nature of the problem statement, but as it turns out that one can figure it exactly, no such simplification is needed. Though on a test, or on a job, both of which have time limits, I'd make it in a heartbeat and only worry about it after checking other, less straightforward work.
i would help, i really wanna but my ap chem class sucked it all outta me
use stoichiometry
NaCl MW= 58.442 g/mole
KCl MW = 74.551 g/mole
AgCl MW= 143.321 g/mole
1 g = weight of NaCl + weight of KCl
say x = weight of NaCl
1-x = weight of KCl
x * mole/58.442g * 143.321 g/mole AgCl = 2.452xg AgCl
(1-x) * mole/74.551g * 143.321 g/mole AgCl = (1-x) 1.922 g AgCl
2.452xg AgCl + (1-x) 1.922g AgCl = 2.1476 g AgCl
2.452xg AgCl + 1.922gAgCl - 1.922xgAgCl = 2.1476 g AgCl
0.53 x g AgCl = 0.2256 g AgCl
x = 0.426 g
Mass percent = mass/ total mass * 100
Mass % NaCl = 0.426g/1g *100 = 43%
KCl = 1-x = 1-0.43 = 0.57
Mass % KCl = 0.57g/1g *100 = 57%
x(grams Cl from NaCl) + (1-x)(grams Cl from KCl) = total grams chloride [not total grams AgCl!]
However, components' formula weights and chlorine wt-% are are different. For relative contributions,
(Cl/NaCl)
(Cl/KCl)
Get the units consistent and wang it all together. Negative roots are not physical. Or, graph the theoretical AgCl yield from one gram of 100% KCl through one gram of 100% NaCl with a few mixtures in-between to demonstrate linearity (or not) and interpolate your answer. Comes to the same thing.
Hey, you top contributor guys, stop doing people's homework problems for them. Feel free to give them tips, but when you do their homework for meaningless points you aren't helping anyone.
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