Question

What is the net ionic equation for the following reaction?
$Ba\left(N{O}_3{)}_2\right(aq)+N{a}_{2}S{O}_4(aq)\to BaS{O}_4(s)+2NaN{O}_3(aq)$

• A. $N{a}^{+}+N{O}_3^{-}\to NaN{O}_3$
• B. $B{a}^{2+}+N{O}_3^{-}+N{a}^{+}+S{O}_4^{2-}\to BaS{O}_4+N{a}^{+}+N{O}_3^{-}$
• C. $B{a}^{2+}+2N{O}_3^{-}+2N{a}^{+}+S{O}_4^{2-}\to BaS{O}_4+2N{a}^{+}+2N{O}_3^{-}$
• D. $B{a}^{2+}+S{O}_4^{2-}\to BaS{O}_4$

Answer 1

Answer: (C)

$B{a}^{2+}+2N{O}_3^{-}+2N{a}^{+}+S{O}_4^{2-}\to BaS{O}_4+2N{a}^{+}+2N{O}_3^{-}$

$Ba(N{O}_3{)}_2\left(aq\right)+N{a}_{2}S{O}_4\left(aq\right)\to BaS{O}_4\left(s\right)+2NaN{O}_3\left(aq\right)$

Only aqueous solutions produce ions. Whereas solid molecules don't produce ions.

$BaS{O}_4$ doesn't produce ions in sold state.

The ionic can would be,

$B{a}^{+2}+2N{O}_3^{-}+2N{a}^{+}+S{O}_4^{2-}\to BaS{O}_4+2N{a}^{+}+N{O}_3^{-}$