# Question #8053f

Jan 8, 2015

The answer is $3.0$ $\text{g}$ of $B a \left(S {O}_{4}\right)$ will form in this reaction.

Your general chemical reaction is

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

Notice that you've got a $1 : 1$ mole ratio between $B a {\left(N {O}_{3}\right)}_{2}$ and $B a S {O}_{4}$; this means that for every mole of $B a {\left(N {O}_{3}\right)}_{2}$ used, 1 mole of solid will be produced.

The number of $B a {\left(N {O}_{3}\right)}_{2}$ moles can be determined using its molarity:

$C = \frac{n}{V} \implies {n}_{B a {\left(N {O}_{3}\right)}_{2}} = C \cdot V = 0.50$ $\text{M} \cdot 25 \cdot {10}^{- 3}$ $\text{L}$

${n}_{B a {\left(N {O}_{3}\right)}_{2}} = 0.013$ $\text{moles}$

This means of course that ${n}_{B a S {O}_{4}} = {n}_{B a {\left(N {O}_{3}\right)}_{2}} = 0.013$ moles of solid will be formed. Knowing that $B a S {O}_{4}$'s molar mass is $233.3$ $\text{g/mol}$, the mass produced will be

${m}_{B a S {O}_{4}} = {n}_{B a S {O}_{4}} \cdot 233.3 \frac{g}{m o l} = 0.013$ $\text{moles} \cdot 233.3 \frac{g}{m o l}$

${m}_{B a S {O}_{4}} = 3.0$ $\text{g}$

The reaction's complete ionic equation is

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

The net ionic equation is

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