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write the products of the following double displacement reactions. ab +…

Question

write the products of the following double displacement reactions. ab + cd → ad + cb

  1. al₂o₃ + ba(oh)₂ →

al^3+ o^2- ba^2+ oh^-
al(oh)₃ + bao

  1. mg₃p₂ + al₂(so₄)₃ →

mg^3+ p^2+ al^2+ so^-
mg^2 p^2- + al^3+ so^-
mg2+ al(so)4

  1. k₂s + ca₃n₂ →

k^2+ s^2- ca^3+ n^2+
k(s)^2 + ca(n)^3
k^2+ s^2- ca^2+ n^3+

  1. ga(no₃)₃ + srf₂ →

ga^3+ no^9+ sr^2+ f^2+
ga(no)3 + srf₂
ga^3+ na^3+ sr^2+ f^-

Explanation:

Step1: Determine the products of double - displacement reaction for \(Al_2O_3+Ba(OH)_2\)

In a double - displacement reaction \(AB + CD
ightarrow AD+CB\). For \(Al_2O_3\) (\(A = Al^{3 +}\), \(B=O^{2 -}\)) and \(Ba(OH)_2\) (\(C = Ba^{2+}\), \(D = OH^{-}\)), we swap the ions.
The charge on \(Al^{3+}\) and \(OH^{-}\) gives \(Al(OH)_3\) (since \(3 + (- 1)\times3=0\)) and the charge on \(Ba^{2+}\) and \(O^{2 -}\) gives \(BaO\) (since \(2+( - 2)=0\)).

Step2: Determine the products of double - displacement reaction for \(Mg_3P_2+Al_2(SO_4)_3\)

For \(Mg_3P_2\) (\(A = Mg^{2+}\), \(B = P^{3 -}\)) and \(Al_2(SO_4)_3\) (\(C=Al^{3+}\), \(D = SO_4^{2 -}\)).
When we swap the ions, for \(Mg^{2+}\) and \(SO_4^{2 -}\), the compound is \(MgSO_4\) (since \(2+( - 2)=0\)) and for \(Al^{3+}\) and \(P^{3 -}\), the compound is \(AlP\) (since \(3+( - 3)=0\)).

Step3: Determine the products of double - displacement reaction for \(K_2S+Ca_3N_2\)

For \(K_2S\) (\(A = K^{+}\), \(B = S^{2 -}\)) and \(Ca_3N_2\) (\(C = Ca^{2+}\), \(D=N^{3 -}\)).
When we swap the ions, for \(K^{+}\) and \(N^{3 -}\), the compound is \(K_3N\) (since \(1\times3+( - 3)=0\)) and for \(Ca^{2+}\) and \(S^{2 -}\), the compound is \(CaS\) (since \(2+( - 2)=0\)).

Step4: Determine the products of double - displacement reaction for \(Ga(NO_3)_3+SrF_2\)

For \(Ga(NO_3)_3\) (\(A = Ga^{3+}\), \(B = NO_3^{-}\)) and \(SrF_2\) (\(C = Sr^{2+}\), \(D = F^{-}\)).
When we swap the ions, for \(Ga^{3+}\) and \(F^{-}\), the compound is \(GaF_3\) (since \(3+( - 1)\times3 = 0\)) and for \(Sr^{2+}\) and \(NO_3^{-}\), the compound is \(Sr(NO_3)_2\) (since \(2+( - 1)\times2=0\)).

Answer:

  1. \(Al_2O_3+Ba(OH)_2

ightarrow 2Al(OH)_3 + BaO\)

  1. \(Mg_3P_2+Al_2(SO_4)_3

ightarrow3MgSO_4 + 2AlP\)

  1. \(K_2S+Ca_3N_2

ightarrow2K_3N+3CaS\)

  1. \(Ga(NO_3)_3+SrF_2

ightarrow GaF_3+Sr(NO_3)_2\)