Question

I did an experiment with a sand sample which contains fine metallic aluminium and I wanted to measure how much of this metallic aluminium is in the sample. I used the NaOH solution in order to let react the aluminium and collect the H2. From the H2 amount I thought I could calculate the amount of aluminium in the sample. However now I do not know how to calculate it, I probably should first calculate the MH2 (mol), then MAl (mol) and from this I could go to Mal (g). Could you help with the equations?

Answer 1

start with the equation for the reaction

2Al + 6H2O --> 2Al3+ + 3H2(g) + 6OH-

This is because the OH- removes the oxide coating from the Al, but the reaction is with the water.

Hence we can see that 2 moles of Al yields 3 moles of H2.

From your volume of H2 (you also need temperature and possibly pressure) you can calculate the number of moles using

PV=nRT

If you do not have the PV information then assume that one mole of H2 occupies 22.4 litres and use this to calculate the number of moles.

The number of moles of Al is then two thirds of the number of moles of H2 (from the equation).

From the number of moles of Al calculate the mass of Al from the molar mass of Al.

Answer 2

Thanks, are these equations correct?

M (H2 in mol)=(V(H2 in ml)/1000)/22.4)

then
M (Al in mol)=2/3V(H2 in mol)

after
M (Al in g) = M (Al in mol) x 27

Answer 3

I am not sure that I follow your symbols.

Let V=volume of H2 (in ml)

Number of moles of H2=
V/(22.4 L mol^-1 x 1000 ml L^-1)

Number of moles of Al =
2xV/22.4 L mol^-1 x 1000 ml L^-1 x3)

mass of Al =
2xVx27 g mol^-1/(22.4 L mol^-1 x1000 ml L^-1 x3)

Answer 4

Thanks a lot.