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?

40.8g of sodium sulfate must be added

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) =

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.