Chemistry: dissolving aluminum metal in hydrochloric acid

In summary, the problem involves finding the volume of 1.58 M HCl needed to dissolve 3.200 g of aluminum. The single-replacement reaction of 2Al(s) + 6HCl(aq) = 2AlCl[3](aq) + 3H[2](g) is used to solve the problem. The stepwise path includes finding the moles of aluminum and HCl, and then using the concentration of the HCl solution to find the volume needed. It is also mentioned that a catalyst may be needed to accelerate the breakdown of surface passivation, but this is likely not applicable for a beginning level course.
  • #1
Madelin Pierce
24
2

Homework Statement


Aluminum metal dissolves in hydrochloric acid. What volume, in mL, of 1.58 M HCl is needed to completely dissolve 3.200 g of aluminum?(Hint: write the single-replacement reaction first)

Homework Equations


V[1]M[1]=V[2]M[2]

The Attempt at a Solution


I got the single-replacement reaction:
2Al(s) + 6HCl(aq) = 2AlCl[3](aq) + 3H[2](g)
But from there, I'm not really sure what to do. I think stoichiometry's involved, but I'm not sure how to use it.
 
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  • #2
If you have
Madelin Pierce said:
2Al(s) + 6HCl(aq) = 2AlCl[3](aq) + 3H[2](g)
you have the stoechiometry ! My guess is that it's time to convert mol(ecul)es to mass :rolleyes:
 
  • #3
BvU said:
My guess is that it's time to convert mol(ecul)es to mass

I would rather convert mass to moles :wink:
 
  • #4
Sure, and perhaps you can explain to the audience why this is a better idea (generally or in this sspecific case) ?
 
  • #5
Stepwise path to find the volume of HCl,

Find moles of Aluminum (need formula weight of Aluminum)
Find moles of HCl (use ratio according to reaction)
Find volume of HCl solution of 1.58 M (concentration computation)
 
  • #6
Am I correct in remembering that you may need to add a near-catalytic amount of copper chloride to accelerate breakdown of surface passivation ?

Or was this just for 'Activated Zinc' ??
 
  • #7
Nik_2213 said:
Am I correct in remembering that you may need to add a near-catalytic amount of copper chloride to accelerate breakdown of surface passivation ?

Or was this just for 'Activated Zinc' ??
Not in range. O.P. wants entry level course help, like for a beginning course. I am not saying that some kind of catalyst is not needed, but just that the idea is not in the range of the person's study.
 

Related to Chemistry: dissolving aluminum metal in hydrochloric acid

1. What happens when aluminum metal is dissolved in hydrochloric acid?

When aluminum metal is dissolved in hydrochloric acid, a chemical reaction takes place. The acid reacts with the aluminum to produce aluminum chloride and hydrogen gas. This reaction is exothermic, meaning it releases heat.

2. What is the chemical equation for dissolving aluminum metal in hydrochloric acid?

The chemical equation for this reaction is: 2Al + 6HCl → 2AlCl3 + 3H2. This means that for every 2 molecules of aluminum, 6 molecules of hydrochloric acid are needed to produce 2 molecules of aluminum chloride and 3 molecules of hydrogen gas.

3. Why does the solution turn cloudy when aluminum metal is added to hydrochloric acid?

The cloudiness in the solution is due to the formation of hydrogen gas bubbles. As the reaction between the aluminum and hydrochloric acid produces hydrogen gas, these bubbles can make the solution appear cloudy. The cloudiness will dissipate as the reaction progresses and the gas is released.

4. Is it safe to handle aluminum metal and hydrochloric acid for this experiment?

It is important to exercise caution when handling both aluminum metal and hydrochloric acid. Aluminum metal can be sharp and may cause cuts or abrasions, while hydrochloric acid is a corrosive substance that can cause burns if it comes into contact with skin. It is recommended to wear protective gloves and eyewear while conducting this experiment.

5. Can other metals be dissolved in hydrochloric acid in a similar way?

Yes, other metals such as zinc, magnesium, and iron can also be dissolved in hydrochloric acid in a similar manner. However, the rate and extent of the reaction may vary depending on the type of metal and concentration of the acid. It is important to research and understand the properties of the metal and acid being used before attempting any experiments.

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