Question

Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams

of sodium sulfate should a chemist use to completely react with 75.0 grams of barium nitrate? How many

grams of precipitate will be formed?

Answer 1

40.8g of sodium sulfate must be added

Step-by-step explanation:

The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:

Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)

That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:

75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄

Mass Na₂SO₄ -Molar mass: 142.04g/mol-:

0.287 moles * (142.04g / mol) =

40.8g of sodium sulfate must be added