Question

Write balanced chemical equations for each of the following verbal descriptions: (a) Vanadium oxytrichloride \(\left(\mathrm{VOCl}_{3}\right)\) is formed by the reaction of vanadium(III) chloride with oxygen. (b) Niobium(V) oxide is reduced to the metal with hydrogen gas. (c) Iron(III) ion in aqueous solution is reduced to iron(II) ion in the presence of zinc dust. (d) Niobium(V) chloride reacts with water to yield crystals of niobic acid \(\left(\mathrm{HNbO}_{3}\right)\).

Short answer

(a) \(2 \mathrm{VCl}_{3} + \frac{3}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{VOCl}_{3}\) (b) \(\mathrm{Nb}_{2}\mathrm{O}_{5} + 5 \mathrm{H}_{2} \rightarrow 2 \mathrm{Nb} + 5 \mathrm{H}_{2}\mathrm{O}\) (c) \(2 \mathrm{Fe}^{3+}_{(aq)} + 3 \mathrm{Zn}_{(s)} \rightarrow 2 \mathrm{Fe}^{2+}_{(aq)} + 3 \mathrm{Zn}^{2+}_{(aq)}\) (d) \(\mathrm{NbCl}_{5} + 5 \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{HNbO}_{3} + 5 \mathrm{HCl}\)

Step-by-step solution

(a) Writing the balanced chemical equation for the formation of Vanadium oxytrichloride

To write the balanced chemical equation for the formation of Vanadium oxytrichloride (VOCl₃), we first need to identify the reactants and products. According to the description, the reaction involves Vanadium(III) chloride and oxygen. The balanced chemical equation is: \(2 \mathrm{VCl}_{3} + \frac{3}{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{VOCl}_{3}\)

(b) Writing the balanced chemical equation for the reduction of Niobium(V) oxide to Niobium metal

To write the balanced chemical equation for the reduction of Niobium(V) oxide to Niobium metal, we first need to identify the reactants and products. According to the description, the reaction involves Niobium(V) oxide and Hydrogen gas. The balanced chemical equation is: \(\mathrm{Nb}_{2}\mathrm{O}_{5} + 5 \mathrm{H}_{2} \rightarrow 2 \mathrm{Nb} + 5 \mathrm{H}_{2}\mathrm{O}\)

(c) Writing the balanced chemical equation for Iron(III) ion reduced to iron(II) ion in the presence of zinc dust

To write the balanced chemical equation for Iron(III) ion reduced to iron(II) ion in the presence of zinc dust, we first need to identify the reactants and products. According to the description, the reaction involves Iron(III) ion, iron(II) ion, and zinc dust. The balanced chemical equation is: \(2 \mathrm{Fe}^{3+}_{(aq)} + 3 \mathrm{Zn}_{(s)} \rightarrow 2 \mathrm{Fe}^{2+}_{(aq)} + 3 \mathrm{Zn}^{2+}_{(aq)}\)

(d) Writing the balanced chemical equation for Niobium(V) chloride reacts with water to yield crystals of niobic acid

To write the balanced chemical equation for Niobium(V) chloride reacts with water to yield crystals of niobic acid (HNbO₃), we first need to identify the reactants and products. According to the description, the reaction involves Niobium(V) chloride and water. The balanced chemical equation is: \(\mathrm{NbCl}_{5} + 5 \mathrm{H}_{2}\mathrm{O} \rightarrow \mathrm{HNbO}_{3} + 5 \mathrm{HCl}\)