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
Dec 14, 2016

Answer:

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.