Lab Report Chem HL: HCl and aluminium

[Alys]

help!

how much hydrochloric acid (in excess) is needed to fully react 1g, 2g, 3g, 4g, and 5g of aluminium (in separate experiments)

thanks guys

[dessskris]

It depends on the concentration!

What's the formula for aluminium? Is it only Al? If so, then:

n(Al) = mass(Al) / Ar(Al)

n(Al) = 5 / 26.98

n(Al) = 0.185 mol

6 HCl + 2 Al --> 2 AlCl₃ + 3 H₂

I hope it's correct btw

n(HCl)= 6/2 * n(Al)

n(HCl)= 3 * 0.185

n(HCl)= 0.555 mol

Now it depends on the concentration of HCl. I would suggest 1 mol dm^-3 btw.

Then assuming [HCl] = 1 mol dm^-3,

V(HCl) = n(HCl) / [HCl]

V(HCl) = 0.555 / 1

V(HCl) = 0.555 dm³

V(HCl) = 555 cm³

So if you want an excess volume of HCl, I would suggest 600 cm³ or higher

[Drake Glau]

I would suggest using a higher molarity honestly. 600mL is a lot to mess around with. Just my opinion though.

For 2moldm^-3 would be 277.5cm³

I know at my school the bottle comes at 6 molar already so 0.555/6=92.5mL which would be easier to handle in my opinion, however much stronger and will need much more care.

[dessskris]

Just a reminder, I don't know if your teacher wants you to put explanations in Design IAs, but that's how we do it and if we were to do this kind of design, we would need to explain why we chose a certain value as the concentration of the acid. FYI just in case if you don't know, 2 mol/dm3 is the maximum concentration allowed as our teacher said because more than 2 mol/dm3 is too concentrated and thus too harmful for you to do an experiment on. I don't know if this is just a suggestion from our teacher or this is the rule set by the IBO, but it's good to know that and you can put that as a reason why you chose that concentration value. Considering what Walter said, I'd advise 1.5 M or 2 M.

[Alys]

It depends on the concentration!

What's the formula for aluminium? Is it only Al? If so, then:

n(Al) = mass(Al) / Ar(Al)

n(Al) = 5 / 26.98

n(Al) = 0.185 mol

6 HCl + 2 Al --> 2 AlCl₃ + 3 H₂

I hope it's correct btw

n(HCl)= 6/2 * n(Al)

n(HCl)= 3 * 0.185

n(HCl)= 0.555 mol

Now it depends on the concentration of HCl. I would suggest 1 mol dm^-3 btw.

Then assuming [HCl] = 1 mol dm^-3,

V(HCl) = n(HCl) / [HCl]

V(HCl) = 0.555 / 1

V(HCl) = 0.555 dm³

V(HCl) = 555 cm³

So if you want an excess volume of HCl, I would suggest 600 cm³ or higher

Thanks very much, that's really useful your formula is right

I would suggest using a higher molarity honestly. 600mL is a lot to mess around with. Just my opinion though.

For 2moldm^-3 would be 277.5cm³

I know at my school the bottle comes at 6 molar already so 0.555/6=92.5mL which would be easier to handle in my opinion, however much stronger and will need much more care.

thanks, i'll check with my teacher to see what we have and go from there

Just a reminder, I don't know if your teacher wants you to put explanations in Design IAs, but that's how we do it and if we were to do this kind of design, we would need to explain why we chose a certain value as the concentration of the acid. FYI just in case if you don't know, 2 mol/dm3 is the maximum concentration allowed as our teacher said because more than 2 mol/dm3 is too concentrated and thus too harmful for you to do an experiment on. I don't know if this is just a suggestion from our teacher or this is the rule set by the IBO, but it's good to know that and you can put that as a reason why you chose that concentration value. Considering what Walter said, I'd advise 1.5 M or 2 M.

Thanks again i spoke to my teacher today and we use 2M solution