HCl(aq)+Li2CO3(aq)→
Express your answer as a chemical equation. Identify all of the phases in your answer. Enter noreaction if no precipitate is formed.
2HCl + Li2CO3 = 2LiCl(aq) + CO2(g) + H2O
BUT
LiCl is in the form Li+ and CL-
no precipitate
so they tell you to say no reaction, although CO2 bubbles off indicating that there was a reaction :) Beats me.
Perhaps I can clear that up. No reaction is not an option in this case.
In simple terms a reaction occurs when
a. a ppt is formed
b. a gas is formed or
c. a weak electrolyte is formed (such as H2O)
So a reaction does take place for both b and c reasons.
Since the question wants phases I would write
2HCl (aq) + Li2CO3(aq) ==> 2LiCl(aq) + CO2(g) + H2O()l)
Other kinds of reactions are redox, etc that may or may not conform to the three simple reasons above.
The balanced chemical equation for the reaction between hydrochloric acid (HCl) and lithium carbonate (Li2CO3) is:
2HCl(aq) + Li2CO3(aq) -> 2LiCl(aq) + H2O(l) + CO2(g)
In this reaction, hydrochloric acid and lithium carbonate react to produce lithium chloride (LiCl), water (H2O), and carbon dioxide (CO2). The phases are indicated as (aq) for aqueous solution, (l) for liquid, and (g) for gas.
To determine the chemical equation and phases of the reaction between HCl(aq) and Li2CO3(aq), we need to consider the solubility of the products.
HCl(aq) is hydrochloric acid, which is a strong acid that completely dissociates in water. It will ionize into H+ (aq) and Cl- (aq).
Li2CO3(aq) is lithium carbonate, which is a water-soluble ionic compound. It will dissociate into Li+ (aq) and CO3^2- (aq) ions.
Now, let's combine the ions and write the balanced chemical equation:
2H+(aq) + CO3^2-(aq) + 2Li+(aq) + 2Cl-(aq) → 2Li+(aq) + 2Cl-(aq) + CO3^2-(aq) + H2O(l)
Both Li+ and Cl- ions appear on both sides of the equation, so they can be canceled out. This leaves us with:
2H+(aq) + CO3^2-(aq) → CO3^2-(aq) + H2O(l)
Since CO3^2- remains in solution, there is no precipitate formed. Therefore, the answer is "noreaction" as no precipitate is formed.