Answer:
There must be present 312 g of barium chloride. (option d)
Explanation:
Ba⁺² + SO4⁻² → BaSO4
BaCl2 → Ba⁺² + 2Cl⁻
Fe2(SO4)3 → 2Fe⁺³ + 3SO4⁻²
Molar mass Fe2(SO4)3 = 400 g/m
Mass / Molar mass = Moles
200g /400 g/m = 0.5 moles
If one mol of Iron(III) sulfate dissociates in 3 moles of sulfate,
how many moles of sulfate do i have, from 0.5 moles of Iron (III) sulfate.
1 mol Fe2(SO4)3 ____ 3 moles of SO4⁻²
0.5 mol Fe2(SO4)3 ____ (0.5 . 3) /1 = 1.5 moles
Then 1 mole of sulfate anion will react with one mole of barium cation, which in turn comes from one mole of barium chloride, then 1.5 moles of sulfate anion, react with 1.5 moles of barium cation, which come from 1.5 moles of chloride. <u><em>The ratio is always 1 to 1.</em></u>
As we have the moles of barium chloride, let's find out the mass, with the molar weight.
Moles . molar weight = mass
1.5 moles . 208g/m = 312 g
