Erica H.

asked • 10/10/20

how many grams of barium sulfate will be formed upon the complete reaction?

According to the following reaction, how many grams of barium sulfate will be formed upon the complete reaction of 27.9 grams of sulfuric acid with excess barium hydroxide?


Ba(OH)2 (aq) + H2SO4 (aq) BaSO4 (s) + 2H2O (l)

Dawn R.

So first you have to convert the grams of sulfuric acid to moles, giving you 0.285 mol sulfuric acid. Then you see that using the molar ratio of 1 mol sulfuric acid giving you 1 mol barium sulfate you end up with 0.285 mol barium sulfate. Convert that back to grams giving you 66.25 g barium sulfate. Does that make sense?
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10/10/20

Robert S.

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I agree with Dawn R.'s solution. Remember that the molar mass is the sum of the atomic weights of all the individual atoms in the compound, so some that are duplicated. like Ba(OH)2 has two oxygen and two hydrogen atoms that need to be added to the total. The unit is grams/mole.
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10/10/20

1 Expert Answer

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J.R. S. answered • 10/10/20

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Ba(OH)2 (aq) + H2SO4 (aq) ==> BaSO4 (s) + 2H2O (l) ... BALANCED EQUATION

Not sure why Dawn R. and Robert S. wrote their answers in the comment sections, as opposed to in the "answer" section. I didn't see the comments so I answered it here.


Using dimensional analysis and the mol ratio of the balanced equation, we have...

27.9 g H2SO4 x 1 mol H2SO4/98 g 1 mol BaSO4/mol H2SO4 x 233 g BaSO4/mol = 66.3 g BaSO4 (3 sig.figs.)

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