If 45.5 mL of 0.150 M sodium sulfate solution reacts completely with aqueous barium nitrate, what is the mass of #BaSO_4# (233.40 g/mol) precipitate? #Ba(NO_3)_2(aq) + Na_2SO4 (aq) -> BaSO_4(s) + 2NaNO_3(aq)#?

To find the mass of the BaSO4 precipitate, first, calculate the moles of Na2SO4 using the given volume and concentration. Then, use the stoichiometry of the reaction to find the moles of BaSO4 formed. Finally, use the molar mass of BaSO4 to find the mass of the precipitate.

1. Calculate moles of Na2SO4: ( \text{Volume} \times \text{Molarity} = 45.5 \text{ mL} \times 0.150 \text{ mol/L} = 6.825 \text{ mmol} )

2. Use stoichiometry to find moles of BaSO4: From the balanced equation, 1 mole of BaSO4 forms for every mole of Na2SO4 reacted. ( \text{Moles of BaSO}_4 = \frac{6.825 \text{ mmol Na}_2\text{SO}_4}{1} = 6.825 \text{ mmol BaSO}_4 )

3. Convert moles of BaSO4 to grams: ( \text{Mass of BaSO}_4 = \text{moles} \times \text{molar mass} = 6.825 \text{ mmol} \times \frac{233.40 \text{ g}}{\text{mol}} = 1594.94 \text{ mg} ) ( \text{Mass of BaSO}_4 = 1.59494 \text{ g} )

So, the mass of BaSO4 precipitate formed is approximately 1.595 grams.