SOLUTION: I know this is not a Algebra question its for Chemistry 16.05 grams of lead 4 phosphate is combined with ammonium carbonate with ammonium carbonate. Find: a)The mass of ammon

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Question 580766: I know this is not a Algebra question its for Chemistry
16.05 grams of lead 4 phosphate is combined with ammonium carbonate with ammonium carbonate. Find:
a)The mass of ammonium carbonate (NH^4)2CO^3 required
b)The mass of Plumbic carbonate produced
Please your help would be most appreciated
Thank you:)
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
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mychemistrytutor.com.
I used to answer questions there, but I was not really needed there. The need for help is much greater here.
The formula weight for %28NH%5B4%5D%29%5B2%5DCO%5B3%5D can be calculated by combining the atomic masses of nitrogen (14.01), hydrogen (1.008), carbon (12.01) and oxygen (16.00).
2%2814.01%2B4%2A1.008%29%2B12.01%2B3%2A16.00=2%2A18.042%2B12.01%2B48.00=96.094
So we say that 1 mol of ammonium carbonate has a mass of 96.1 grams
Pb%5B3%5D%28PO%5B4%5D%29%5B4%5D would be the formula for lead IV phosphate
Its formula weight would be found by adding the proper combination of the atomic masses of lead (207.2), phosphorus (30.97) and oxygen (16.00)

So we say that 1 mol of lead IV phosphate has a mass of 1001.5 grams.
Pb%28CO%5B3%5D%29%5B2%5D would be the formula for lead IV carbonate.
We can calculate the formula weight as
207.2%2B2%2812.01%2B3%2A16.00%29=207.2%2B2%2A%2860.01%29=207.2%2B120.02=327.22
So we say that 1 mol of lead IV carbonate has a mass of 327.2 grams.
Looking at the formulas we figure that the balanced reaction must be
Pb%5B3%5D%28PO%5B4%5D%29%5B4%5D%2B6%28NH%5B4%5D%29%5B2%5DCO%5B3%5D=3Pb%28CO%5B3%5D%29%5B2%5D+4%28NH%5B4%5D%29%5B3%5DPO%5B4%5D
meaning that each mole of lead IV phosphate will react with 6 moles ammonium carbonate to yield 3 moles of lead IV carbonate.
So, the mass of ammonium carbonate needed is
(16.05 g Pb phosphate)(1 mol Pb phosphate/1001.5 g Pb phosphate)(6 mol NH4 carbonate/1 mol Pb phosphate)(96.1 g NH4 carbonate/1 mol NH4 carbonate = 9.24 g NH4 carbonate
16.05%2A%281%2F1001.5%29%286%2F1%29%2896.1%2F1%29=9.24
The mass of Pb IV carbonate produced would be
(16.05 g Pb phosphate)(1 mol Pb phosphate/1001.5 g Pb phosphate)(3 mol Pb carbonate/1 mol Pb phosphate)(327.2 g Pb carbonate/1 mol Pb carbonate = 15.73 g Pb carbonate
16.05%2A%281%2F1001.5%29%283%2F1%29%28327.2%2F1%29=15.73