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##