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

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

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Answer 1 · 2016-12-14T08:45:39

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

Explanation:

$B a C l_{2} (a q) + N a_{2} S O_{4} (a q) \to B a S O_{4} (s) ⏐ ↓ + 2 N a C l (a q)$ $BaCl_2(aq) + Na_2SO_4(aq) rarr BaSO_4(s)darr + 2NaCl(aq)$

$Moles of barium chloride = \frac{18.48 \cdot g}{208.23 \cdot g \cdot m o l^{- 1}} = 8.87 \times 10^{- 2} \cdot m o l$ $"Moles of barium chloride"=(18.48*g)/(208.23*g*mol^-1)=8.87xx10^-2*mol$

Barium chloride is the limiting reagent. At MOST, we can get $8.87 \times 10^{- 2} \cdot m o l$ $8.87xx10^-2*mol$ $B a S O_{4} (s)$ $BaSO_4(s)$ ,

i.e. $8.87 \times 10^{- 2} \cdot m o l \times 233.38 \cdot g \cdot m o l^{- 1} = ? ? g$ $8.87xx10^-2*molxx233.38*g*mol^-1=??g$

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

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

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