A student observed the following reaction:
AlCl3(aq)+3NaOH(aq) -> Al(OH)3(s)+3NaCl(aq)
After the products were filtered, which substances remained on the filter paper and why?
Just because you know it doesn’t mean other people don’t smh
A10H3 s
at least answer the question like what are you doing
To determine which substances remained on the filter paper after the reaction, we need to understand the properties of the reactants and the products.
In the given reaction, AlCl3 and NaOH are the reactants, while Al(OH)3 and NaCl are the products.
AlCl3 is a soluble compound, meaning it dissolves completely in water. When it is mixed with water, it dissociates into Al3+ and Cl- ions.
NaOH is also a soluble compound that dissociates in water, forming Na+ and OH- ions.
Now, when AlCl3 reacts with NaOH, they undergo a double displacement reaction. The Al3+ ion from AlCl3 combines with the OH- ion from NaOH to form Al(OH)3, which is insoluble in water and precipitates out as a solid.
On the other hand, the Na+ and Cl- ions from both AlCl3 and NaOH remain in the solution as they are both soluble, forming NaCl.
To separate the solid Al(OH)3 from the remaining solution, the student filters the mixture. The filter paper acts as a barrier that allows liquid (the solution) to pass through but holds back the solid particles. As a result, the solid Al(OH)3 remains on the filter paper, while the dissolved NaCl passes through with the filtrate.
Therefore, after the products were filtered, the substance that remains on the filter paper is Al(OH)3, while the substance that passes through is NaCl.