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What is the composition of the salt mixture?
2Na3PO4 + 3BaCl2 ----> Ba3(PO4)2 + NaCl
The precipitate was found to be Ba 2+
The mass of the unknown salt mixture was 0.9900 g
Mass of Ba3(PO4)2 precipitate: 0,2691 g
So to calculate the composition I have to find the actual yield and divide it by theoretical yield right?
Since the actual yield is determined experimentally 0.2691 g, how do I find theoretical.
Do I used the 0,9900 g?
1 Answer
- Trevor HLv 79 years agoFavorite Answer
This is a very unclear question.I can only guess that you have a mixture of Na3PO4 and BaCl2 with mass 0.9900g. You dissolved this mixture in water and a precipitate was produced
You have correctly written an equation for this reaction:
You precipitated 0.2691g Ba3(PO4)2
How many grams of BaCl2 produced this?
Molar mass Ba3(PO4)2 = 601.9 g/mol
Molar mass BaCl2 = 208.2g/mol - 3BaCl2 = 624.6g
0.2691g Ba3(PO4)2 comes from 0.2691 * 624.6/601.9 = 0.2792g BaCl2
If you started with 0.9900g mixture:
mass of BaCl2 = 0.2792g
Mass of Na3PO4 = 0.9900 - 0.2792 = 0.7108g
Because I do not understand the actual question - the above is only the best guess I could make.
