How can I solve for this ? Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4.

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La0damDoglothagj La0damDoglothagj

15-06-2016
Chemistry

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How can I solve for this ? Calculate the mass of the precipitate formed when 2.27 L of 0.0820 M Ba(OH)2 are mixed with 3.06 L of 0.0664 M Na2SO4.

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taskmasters taskmasters · Answer 1 · 2016-06-29T04:54:58+02:00

The balanced chemical reaction is:

Na2SO4 + Ba(OH)2 = 2NaOH +BaSO4

We are given the amount of the reactants to be used. These values will be the starting point of our calculations.

0.0820 mol/L Ba(OH)2 (2.27 L solution) = 0.18614 mol Ba(OH)2
0.0664 mol/L Na2SO4 (3.06 L solution) = 0.20318 mol Na2SO4

The limiting reactant is the Ba(OH)2. The amount for this compound will be used.

0.18614 mol Ba(OH)2 (1 mol BaSO4 / 1 mol Ba(OH)2 ) (233.43 g BaSO4 / 1 mol BaSO4)= 43.4507 g BaSO4