Question

1
what are the products of this reaction? *
(0.5 points)
__na₃po₄(aq) + __agno₃(aq) →
○ na₃no₃ + agpo₄
○ nano3 + ag₃po₄
○ na₃n + agpo₅
○ nano₃ + agpo₄

Explanation:

Step1: Identify Reaction Type

This is a double - displacement reaction. In a double - displacement reaction, the general form is \(AB + CD
ightarrow AD+CB\), where the cations and anions of the two reactants switch places. For the reaction between \(Na_{3}PO_{4}(aq)\) and \(AgNO_{3}(aq)\), the cation of \(Na_{3}PO_{4}\) is \(Na^{+}\) and the anion is \(PO_{4}^{3 - }\), and the cation of \(AgNO_{3}\) is \(Ag^{+}\) and the anion is \(NO_{3}^{-}\).

Step2: Determine Products by Ion Switching

When we switch the cations and anions, the new compounds formed will be from \(Na^{+}\) combining with \(NO_{3}^{-}\) and \(Ag^{+}\) combining with \(PO_{4}^{3 - }\).

• The formula for the compound formed from \(Na^{+}\) and \(NO_{3}^{-}\): Since the charge of \(Na^{+}\) is \( + 1\) and the charge of \(NO_{3}^{-}\) is \(- 1\), the formula is \(NaNO_{3}\) (because \(1\times( + 1)+1\times( - 1)=0\)).
• The formula for the compound formed from \(Ag^{+}\) and \(PO_{4}^{3 - }\): Let the number of \(Ag^{+}\) ions be \(x\) and the number of \(PO_{4}^{3 - }\) ions be \(y\). We have \(x\times( + 1)+y\times( - 3)=0\), which gives \(x = 3y\). When \(y = 1\), \(x = 3\), so the formula is \(Ag_{3}PO_{4}\).

Answer:

\(NaNO3 + Ag_{3}PO_{4}\) (the second option: \(NaNO3 + Ag_{3}PO_{4}\))