Chemistry

Hi, i'm in year 12 and i'm doing chemistry.
I have to do this by Monday and I really don't get it.

In the following reactions calculate the mass of precipitate formed from 20 g of the metal salt in each case.

(ii) Al2(SO4)(aq) + 6NaOH → 2 Al(OH)3(s) + 3Na2SO4(aq)
(iii) MgSO4(aq) + 2NaOH → Mg(OH)2(s) + Na2SO4(aq)

Thanks in advance.

Reply 1

oxymoron

The Chameleon

In the following reactions calculate the mass of precipitate formed from 20 g of the metal salt in each case.

(ii) Al2(SO4)(aq) + 6NaOH → 2 Al(OH)3(s) + 3Na2SO4(aq)
(iii) MgSO4(aq) + 2NaOH → Mg(OH)2(s) + Na2SO4(aq)

Don't know where (i) is! but anyway....

ii) Moles Al2(SO4)3 (I'm guessing you missed out the 3 there) in 20g = 20/342 = 0.05848 (because molecular mass of metal salt is 342)

Therefore moles Al(OH)3 produced = 0.05848 x 2 = 0.11696 (from equation)

Therefore mass produced = 0.11696 x 78 = 9.12 g (because molecular mass of precip is 78)

iii) Moles salt = 20/120 = 0.1667
Moles precip = 0.1667
mass pecip = 0.1667 x 58 = 9.67 g