# 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  __ BaSO4 + __ NaNO3

Dec 14, 2014

The answer is $2.79 g$.

Starting from the balanced chemical equation

$B a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + N {a}_{2} S {O}_{4 \left(a q\right)} \to B a S {O}_{4 \left(s\right)} + 2 N a N {O}_{3 \left(a q\right)}$

Taking into consideration the solubility rules (more here: http://www.chem.sc.edu/faculty/morgan/resources/solubility/), we can see that $B a {\left(N {O}_{3}\right)}_{2}$, $N {a}_{2} S {O}_{4}$, and $N a N {O}_{3}$ will dissociate into their respective ions, which will lead to the reaction's net ionic equation

$B {a}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -} \to B a S {O}_{4 \left(s\right)}$

Since sulfate compounds formed with $B {a}^{2 +}$ cations are insoluble in water (only slightly soluble, $B a S {O}_{4}$'s ${K}_{s p}$ being equal to $1.1 \cdot {10}^{- 10}$ ), this double-replacement reaction forms a precipitate, $B a S {O}_{4}$.

We know from the balanced chemical equation that the mole-to-mole ratio of $B a {\left(N {O}_{3}\right)}_{2}$ and $B a S {O}_{4}$ 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 molarity, $C = \frac{n}{V}$

${n}_{B a {\left(N {O}_{3}\right)}_{2}} = C \cdot V = 0.500 M \cdot 25.0 \cdot {10}^{- 3} L = 0.012$

Knowing barium sulfate's molar mass ($233.3 \frac{g}{m o l}$), and the number of moles produced, we get

${m}_{B a S {O}_{4}} = {n}_{B a S {O}_{4}} \cdot m o l a r m a s s = 0.012 m o l e s \cdot 233.3 \frac{g}{m o l e} = 2.79 g$