Question

the spectator ions in the reaction shown are
2 agno₃ (aq) + k₂so₄ (aq) → 2 kno₃ (aq) + ag₂so₄ (s)
potassium ion and sulfate ion.
hydrogen ion and hydroxide ion.
potassium ion and nitrate ion.
silver ion and sulfate ion.
silver ion and nitrate ion.
question 2
1 pts
the balanced molecular equation for complete neutralization of h₂so₄ by koh
in aqueous solution is
h₂so₄ (aq) + 2koh (aq) → 2h₂o (l) + k₂so₄ (aq)
h₂so₄ (aq) + 2koh (aq) → 2h₂o (l) + 2k₂so₄ (aq)
2h₂so₄ (aq) + 2koh (aq) → 2h₂o (l) + k₂so₄ (aq)
2h₂so₄ (aq) + 2koh (aq) → h₂o (l) + 2k₂so₄ (aq)

Explanation:

Question 1

To find spectator ions, first write the total ionic equation.
For \(2AgNO_3(aq)+K_2SO_4(aq)\to2KNO_3(aq)+Ag_2SO_4(s)\), the total ionic equation is \(2Ag^{+}(aq)+2NO_3^{-}(aq)+2K^{+}(aq)+SO_4^{2 -}(aq)\to2K^{+}(aq)+2NO_3^{-}(aq)+Ag_2SO_4(s)\).
Spectator ions are ions that appear on both sides of the ionic equation. Here, \(K^{+}\) (potassium ion) and \(NO_3^{-}\) (nitrate ion) are on both sides.

For the neutralization reaction of \(H_2SO_4\) (a diprotic acid) and \(KOH\) (a monobasic base).
The general form of acid - base neutralization is \(H_2SO_4 + KOH\to K_2SO_4+H_2O\).
Using the principle of conservation of atoms (balance the number of each type of atom on both sides of the equation).
For hydrogen atoms: In \(H_2SO_4\), there are 2 \(H\) atoms. In \(KOH\), there is 1 \(H\) atom. In \(H_2O\), there are 2 \(H\) atoms. For sulfate \((SO_4^{2-})\) and potassium \((K^{+})\) ions, we find that \(H_2SO_4(aq)+2KOH(aq)\to2H_2O(l)+K_2SO_4(aq)\) balances the equation. Count the number of \(S\): 1 on both sides, \(O\): \(4 + 2=6\) (left - hand side from \(H_2SO_4\) and \(KOH\)) and \(2 + 4 = 6\) (right - hand side from \(H_2O\) and \(K_2SO_4\)), \(K\): 2 on both sides, \(H\): \(2+2 = 4\) (left - hand side) and \(2\times2=4\) (right - hand side).

Answer:

potassium ion and nitrate ion.

Question 2