Question

write the balanced molecular chemical equation for the reaction in aqueous solution for sodium hydroxide and tin(iv) acetate. if no reaction occurs, simply write only nr. be sure to include the proper phases for all species within the reaction.

Explanation:

Step1: Write the reactants

Sodium hydroxide (\(NaOH\)) is aqueous (\(aq\)), tin(IV) acetate (\(Sn(C_2H_3O_2)_4\)) is aqueous (\(aq\)).
$$NaOH(aq)+Sn(C_2H_3O_2)_4(aq)$$

Step2: Predict the products using double - displacement reaction

In a double - displacement reaction \(AB + CD
ightarrow AD+CB\). Here \(A = Na^+\), \(B=OH^-\), \(C = Sn^{4 +}\), \(D = C_2H_3O_2^-\). The products are sodium acetate (\(NaC_2H_3O_2\)) and tin(IV) hydroxide (\(Sn(OH)_4\)). Sodium acetate is soluble in water (\(aq\)), tin(IV) hydroxide is insoluble (\(s\)).

$$NaOH(aq)+Sn(C_2H_3O_2)_4(aq) ightarrow NaC_2H_3O_2(aq)+Sn(OH)_4(s)$$

Step3: Balance the equation

For sodium (\(Na\)): There is \(1\) \(Na\) on the left and \(1\) on the right. For tin (\(Sn\)): \(1\) on both sides. For acetate (\(C_2H_3O_2\)): \(4\) on the left (from \(Sn(C_2H_3O_2)_4\)) and \(1\) on the right (from \(NaC_2H_3O_2\)). Multiply \(NaC_2H_3O_2\) by \(4\).

$$NaOH(aq)+Sn(C_2H_3O_2)_4(aq) ightarrow 4NaC_2H_3O_2(aq)+Sn(OH)_4(s)$$

Now for \(Na\): we have \(4\) on the right. Multiply \(NaOH\) by \(4\).

$$4NaOH(aq)+Sn(C_2H_3O_2)_4(aq) ightarrow 4NaC_2H_3O_2(aq)+Sn(OH)_4(s)$$

Check the balance of \(O\) and \(H\):

• For \(O\): In \(4NaOH\): \(4\) \(O\) from \(OH^-\). In \(Sn(OH)_4\): \(4\) \(O\) from \(OH^-\). Total \(O\) from hydroxide: \(4 + 4=8\). In \(Sn(C_2H_3O_2)_4\): \(8\) \(O\) from acetate (\(4\times2\)). In \(4NaC_2H_3O_2\): \(8\) \(O\) from acetate (\(4\times2\)).
• For \(H\): In \(4NaOH\): \(4\) \(H\) from \(OH^-\). In \(Sn(OH)_4\): \(4\) \(H\) from \(OH^-\). Total \(H\) from hydroxide: \(4 + 4 = 8\).

Answer:

$$4NaOH(aq)+Sn(C_2H_3O_2)_4(aq) ightarrow 4NaC_2H_3O_2(aq)+Sn(OH)_4(s)$$