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Limiting Reactant

  • Date Submitted: 03/21/2010 04:50 PM
  • Flesch-Kincaid Score: 74.8 
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The objective of this lab was to determine how much precipitate would be formed if BaCl2   was reacted with (NH4)2CO3.
BaCl2 + (NH4)2CO3 ----> BaCO3 + 2(NH4)Cl


To perform this lab, you will need;
Ring stand
2 pieces of filter paper
4 empty beakers
4.5g of (NH4)2CO3
7g of BaCl2
Funnel, small enough to fit in ring stand.
2 small test tubes

Begin by pouring the 7g of BaCl2 and the 4.5g of (NH4)2CO3 into two seperate beakers with equal amounts of water. Because both of these compounds are soluble, they should dissolve in the water. Next, mix the two aqueous solutions together in your third beaker. A reaction should take place almost immediately. The reaction taking place is known as double displacement, meaning that the two solutions trade partners. This leaves you with BaCO3 and 2(NH4)Cl. Based on the 4 solubility rules, you can determine that the precipitate formed is BaCO3.
Next, put the folded piece of filter paper into the funnel on the ring stand. Place the 4th beaker under the funnel, and slowly pour in the final solution. The precipitate, BaCO3, will stay on the filter paper, while the 2(NH4)Cl drips through. This is known as the supernatent. Next, test the supernatent for excess 2(NH4)Cl. Because you started with more (NH4)2CO3, the supernatent should contain excess   2(NH4)Cl. To do this, take 2 small samples of the supernatent in small test tubes. Next, pour in the original 2 solutions. As the (NH4)2CO3 is poured in, there should be no change, simply because a solution cannot react with itself. As BaCl2 is poured in, there should be a mild reaction that turns the solution a milky white. This means that there was excess (NH4)2CO3, and that the solution is filtering correctly. Once the first piece of filter paper is full, take it out, and put in the second. After the solution has been completely filtered, the filter paper must be dried. Once the filter paper is completely dry, it can...


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