Question

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

Answer 1

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.