What is the identity of the precipitate that forms when 70 mL of 0.3 M BaCl2 reacts with 65 mL of 0.35 M Na2CO3
barium carbonate
in the same family as calcium carbonate (chalk)
Oh, you're asking me about precipitates? Well, let me tell you, they're like that uninvited guest at a party. You don't really want them there, but they show up anyway and cause all sorts of trouble. But hey, let's get down to business.
So, when you mix BaCl2 and Na2CO3, a reaction occurs and a precipitate is formed. In this case, we're trying to figure out the identity of that pesky precipitate.
First, we need to find out which ion from each compound will precipitate out. BaCl2 dissociates into Ba2+ ions and 2 Cl- ions, while Na2CO3 dissociates into 2 Na+ ions and CO32- ions. Now, let's see if any of these ions can get together and form a solid.
Ba2+ ions can combine with CO32- ions to form BaCO3, which is insoluble in water and thus precipitates out. Voila! The identity of the precipitate is none other than good old Barium Carbonate (BaCO3).
So, next time you stumble upon a precipitate, just remember that it's like the party crasher of chemical reactions. Hope that clears things up!
To determine the identity of the precipitate formed when 70 mL of 0.3 M BaCl2 reacts with 65 mL of 0.35 M Na2CO3, we need to consider the stoichiometry of the reaction and the solubility rules.
The balanced chemical equation for the reaction between BaCl2 and Na2CO3 is:
BaCl2 + Na2CO3 -> BaCO3 + 2NaCl
From the balanced equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2CO3 to produce 1 mole of BaCO3.
Let's calculate the number of moles of BaCl2 and Na2CO3 used in this reaction:
Moles of BaCl2 = volume (L) x concentration (M) = 0.070 L x 0.3 M = 0.021 moles
Moles of Na2CO3 = volume (L) x concentration (M) = 0.065 L x 0.35 M = 0.023 moles
According to the stoichiometry of the reaction, the molar ratio between BaCl2 and BaCO3 is 1:1. Therefore, the moles of BaCO3 formed in the reaction are also 0.021 moles.
Now, let's use the solubility rules to determine if BaCO3 is soluble or insoluble:
According to the solubility rules, carbonates are generally insoluble except for Group 1 (alkali metal) carbonates and ammonium carbonate.
Therefore, BaCO3 is insoluble and forms a precipitate in the reaction.
In conclusion, the precipitate that forms when 70 mL of 0.3 M BaCl2 reacts with 65 mL of 0.35 M Na2CO3 is BaCO3.
To determine the identity of the precipitate formed when BaCl2 reacts with Na2CO3, we need to consider the reaction between the two compounds. The balanced chemical equation for the reaction between BaCl2 and Na2CO3 is:
BaCl2 + Na2CO3 -> BaCO3 + 2NaCl
From the balanced chemical equation, we can see that when BaCl2 reacts with Na2CO3, it forms BaCO3 as the precipitate.
Now, let's calculate the moles of BaCl2 and Na2CO3 to determine which reactant is limiting and, thus, the amount of BaCO3 that will be formed.
Moles of BaCl2 = Volume (in L) x Molarity
= 0.07 L x 0.3 mol/L
= 0.021 mol
Moles of Na2CO3 = Volume (in L) x Molarity
= 0.065 L x 0.35 mol/L
= 0.02275 mol
According to the balanced chemical equation, BaCl2 and Na2CO3 react in a 1:1 ratio. However, from the above calculations, we can see that the moles of Na2CO3 (0.02275 mol) are slightly greater than the moles of BaCl2 (0.021 mol).
Since Na2CO3 is present in excess, it will completely react with all of the BaCl2. Therefore, the limiting reactant in this reaction is BaCl2.
Now, to find the amount of BaCO3 formed, we use the stoichiometry from the balanced equation:
From the balanced equation, we see that 1 mole of BaCl2 reacts to form 1 mole of BaCO3.
Hence, 0.021 mol of BaCO3 will be formed in this reaction.
Therefore, the identity of the precipitate formed is BaCO3.