Can anyone help me with this? How many moles of aluminium ions and sulfate ions are there when 34.2g of aluminium sulfate is dissolved in 500mL of water?
First find the number of moles of aluminium sulfate using n=m/MM
n(Al2SO43)=34.2/342.15
n(Al2SO43)=0.1
Now find the ratio of the ions in the aluminium sulfate (which is 2Al^3+: 3(SO4)^2-) and multiply these ratios by the original number of moles of aluminium sulfate.
Therefore the number of moles of Al^3+=2x0.1 mol aluminium ions=0.2 mol
number of moles of (SO4)^2-=3x0.1 mol sulfate ions=0.3 mol
@milkycat because there are the ions are only separated in the aqueous solution, mabye a follow up/extension question could be "calculate the concentration of the solution?"
Can you guys help me with this?
Write down the reaction between sodium and sulfuric acid. The solution is then evaporated with a bunsen burner. He then weighs the salt left behind. Calculate expected mass of salt if reaction was complete.
2Na(s) + H2SO4 (aq) -> Na2SO4(aq) + H2(g)
n(Na) = 4 /22.99
= 0.1739 moles
n(Na2SO4) = 0.08695 moles (ratios)
m(Na2SO4) = 0.08695 x (2(22.99) + 32.07 +4(16))
= 12.35g
Is this correct?