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QUESTION

# How many grams of solid BaSO4 will form when Na2SO4 reacts with 25 mL of 0.50 M Ba(NO3)2? __ Ba(NO3)2 + __ Na2SO4 _x0001_ __ BaSO4 + __ NaNO3

The answer is 2.79g.

Starting from the balanced chemical equation

Ba(NO_3)_(2(aq)) + Na_2SO_(4(aq)) -> BaSO_(4(s)) + 2NaNO_(3(aq))

Taking into consideration the solubility rules (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/), we can see that Ba(NO_3)_2, Na_2SO_4, and NaNO_3 will dissociate into their respective ions, which will lead to the reaction's net ionic equation

Ba_((aq))^(2+) + SO_(4(aq))^(2-) -> BaSO_(4(s))

Since sulfate formed with Ba^(2+) cations are insoluble in water (only slightly soluble, BaSO_4's being equal to 1.1 * 10^(-10) ), this forms a precipitate, BaSO_4.

We know from the balanced chemical equation that -to-mole ratio of Ba(NO_3)_2 and BaSO_4 is 1:1; that is, for every mole of barium nitrate used, one mole of barium sulfate is produced.

The number of barium nitrate moles can be determined from its , C = n/V

n_(Ba(NO_3)_2) = C * V = 0.500 M * 25.0 * 10^(-3) L = 0.012

Knowing barium sulfate's molar mass (233.3 g/(mol)), and the number of moles produced, we get

m_(BaSO_4) = n_(BaSO_4) * molarmass = 0.012 mol es * 233.3g/(mol e) = 2.79g