How many grams of barium sulfate solid are produced from reacting 18.48 grams of barium chloride with an excess amount of sodium sulfate?

1 Answer
anor277
Dec 14, 2016

We need, a stoichiometrically balanced equation........and we get approx. 20*g solid sulfate.

Explanation:

BaCl_2(aq) + Na_2SO_4(aq) rarr BaSO_4(s)darr + 2NaCl(aq)

"Moles of barium chloride"=(18.48*g)/(208.23*g*mol^-1)=8.87xx10^-2*mol

Barium chloride is the limiting reagent. At MOST, we can get 8.87xx10^-2*mol BaSO_4(s),

i.e. 8.87xx10^-2*molxx233.38*g*mol^-1=??g

"Barium sulfate" is as soluble in water as a brick, which is why we used sulfate ion to precipitate the salt.

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