Question

You mix 25.0ml of 6.0m LiOH with 75.0ml of MgSO4. Magnesium hydroxide precipitates from the solution.

A. Write a balanced equation for the reaction.

B. What concentration of MgSO4 is required to react with all the LiOH?

C. What mass of magnesium hydroxide is produced?

D. What is the final concentration of Li2SO4(aq)?

Answer 1

A.         2LiOH   + MgSO₄   ---> Mg(OH)₂ + Li₂SO₄

B . As we see above, 2 moles of LiOH react with 1 moles of MgSO4. So, let's first compute the moles of LiOH for our case:

moles of LiOH = V * M = 0.025 L * 6.0 M = 0.15 moles

Now we use the stoichiometric relation:

2 moles LiOH ---- 1 mole of MgSO4

0.15 moles LiOH ----- x?

x = (0.15 moles LiOH * 1 mole MgSO4) / 2 moles LiOH = 0.075 moles MgSO4

Now, let's find the needed concentration of MgSO4. We'll use the MgSO4 volume in this case because this is what they are asking for:

Concentration required = moles / volume = 0.075 moles MgSO4 / 0.075 L = 1 M

C. Again, use stoichiometrics...

2 moles of LiOH ----- 1 mole of Mg(OH)2

0.15 moles LiOH ------ y?

y = (0.15 moles LiOH * 1 mole Mg(OH)2) / 2 moles LiOH = 0.075 moles Mg(OH)2

Please, notice that if we had done this with MgSO4 moles we would have got the same value.

D. The same amount Mg(OH)2 moles is produced for LiSO4, so:

moles LiSO4 = 0.075 moles.

We need the final volume in this case:

Total volume = 25 ml + 75 ml = 100 ml or 0.1 L

Final concentration = 0.075 moles / 0.1 L = 0.75 M.