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

Dec 14, 2016

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

#### Explanation:

$B a C {l}_{2} \left(a q\right) + N {a}_{2} S {O}_{4} \left(a q\right) \rightarrow B a S {O}_{4} \left(s\right) \downarrow + 2 N a C l \left(a q\right)$

$\text{Moles of barium chloride} = \frac{18.48 \cdot g}{208.23 \cdot g \cdot m o {l}^{-} 1} = 8.87 \times {10}^{-} 2 \cdot m o l$

Barium chloride is the limiting reagent. At MOST, we can get $8.87 \times {10}^{-} 2 \cdot m o l$ $B a S {O}_{4} \left(s\right)$,

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

$\text{Barium sulfate}$ is as soluble in water as a brick, which is why we used sulfate ion to precipitate the salt.