Given the net ion equationba2+ + so2−> baso4 ,how many grams of barium chloride must be present to react with 200grams of iron(iii)sulfate? (mwfe2(so4)3 = 400 g/mol, mwbacl2=208 g/mol)a. 564 g. b. 104 g. c. 34.7 g. d. 312 g. e. 1.5 g.

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. The ratio is always 1 to 1.

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

i'm assuming that it's c.) red bank..

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