Question

Complete and balance the following equations for reactions taking place in an acidic solution.

(a)${\mathbf{\text{VO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{+}}}\left(\text{aq}\right){\mathbf{\text{+ SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{VO}}}^{\mathbf{\text{2 +}}}{\mathbf{\text{(aq) + SO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 -}}}\mathbf{\text{(aq)}}$

(b)${\mathbf{\text{Br}}}_{\mathbf{\text{2}}}{\mathbf{\text{(l) + SO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}\mathbf{\to }{\mathbf{\text{Br}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq) + SO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{2 -}}}\mathbf{\text{(aq)}}$

(c)${\mathbf{\text{Cr}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{7}}}^{\mathbf{\text{2 -}}}{\mathbf{\text{(aq) + Np}}}^{\mathbf{\text{4 +}}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{Cr}}}^{\mathbf{\text{3 +}}}{\mathbf{\text{(aq) + NpO}}}_{\mathbf{\text{2}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)}}$

(d)${\mathbf{\text{HCOOH(aq) + MNO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}{\mathbf{\text{(g) + Mn}}}^{\mathbf{\text{2 +}}}\mathbf{\text{(aq)}}$

(e)${\mathbf{\text{Hg}}}_{\mathbf{\text{2}}}{\mathbf{\text{HPO}}}_{\mathbf{\text{4}}}{\mathbf{\text{(s) + Au(s) + Cl}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}\mathbf{\to }{\mathbf{\text{Hg(l) +H}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}{\mathbf{\text{(aq) + AuCl}}}_{\mathbf{\text{4}}}^{\mathbf{\text{-}}}\mathbf{\text{(aq)}}$

Short answer

(a)${\text{2VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right){\text{+ SO}}_{\text{2}}\text{(g)}\to {\text{2VO}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

(b)${\text{Br}}_{\text{2}}{\text{(l) + SO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\to {\text{2Br}}^{\text{-}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}{\text{(aq) + 4H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$

(c)role="math" localid="1663577641072" ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 3Np}}^{\text{4 +}}{\text{(aq) + 2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{2Cr}}^{\text{3 +}}{\text{(aq) + 3NpO}}_{\text{2}}^{\text{2 +}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

(d)role="math" localid="1663577653296" ${\text{5HCOOH(aq) + 2MNO}}_{\text{4}}^{\text{-}}{\text{(aq) + 6H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{5CO}}_{\text{2}}{\text{(g) + 2Mn}}^{\text{2 +}}{\text{(aq) + 14H}}_{\text{2}}\text{O(l)}$

(e)${\text{3Hg}}_{\text{2}}{\text{HPO}}_{\text{4}}{\text{(s) + 2Au(s) + 8Cl}}^{\text{-}}{\text{(aq) + 3H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{6Hg(l) + 3H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq) + 2AuCl}}_{\text{4}}^{\text{-}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Step-by-step solution

What is meant by a balanced chemical equation?

Combining stoichiometric coefficients in the reactants and products is necessary to balance chemical equations. This is significant because a chemical equation must adhere to the laws of conservation of mass and constant proportions, meaning that both the reactant and product sides of the equation must include the same amount of atoms of each element.

Step 2:Balancing equations a, b and c by multiplying equal values.

(a) ${\text{VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right){\text{+ SO}}_{\text{2}}\text{(g)}\to {\text{VO}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Step 1: Balance oxygen atoms.

In this equation, we need to balance oxygen molecules. For these, we can multiply VO by two, and we also need to change a product. The balanced equation is as follows.

${\text{2VO}}_{\text{2}}^{\text{+}}\left(\text{aq}\right){\text{+ SO}}_{\text{2}}\text{(g)}\to {\text{2VO}}^{\text{2 +}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

(b) ${\text{Br}}_{\text{2}}{\text{(l) + SO}}_{\text{2}}\text{(g)}\to {\text{Br}}^{\text{-}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Step1: Balance oxygen atoms.

In this equation, we need to balance the bromine and oxygen atom. In the reactant, two bromine are present, and only one is present in the product. Two oxygen atoms are present in the reactant and four in the product. To balance this equation, we need to add six water molecules.

${\text{Br}}_{\text{2}}{\text{(l) + SO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\to {\text{2Br}}^{\text{-}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}\text{(aq)}$

Step2: Balance hydrogen atoms.

After the addition of six water molecules in the reactant, we need to balance hydrogen atoms. To balance the hydrogen atoms, we need to write three hydronium ions.

${\text{Br}}_{\text{2}}{\text{(l) + SO}}_{\text{2}}{\text{(g) + 6H}}_{\text{2}}\text{O}\to {\text{2Br}}^{\text{-}}{\text{(aq) + SO}}_{\text{4}}^{\text{2 -}}{\text{(aq) + 4H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}$

(c) ${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + Np}}^{\text{4 +}}\text{(aq)}\to {\text{Cr}}^{\text{3 +}}{\text{(aq) + NpO}}_{\text{2}}^{\text{2 +}}\text{(aq)}$

Step 1: Balance oxygen atom.

In the reactant, seven oxygen atoms are present; in the product, only two are present. To balance this, we can add twice hydronium ions into it and balance the equation.

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 3Np}}^{\text{4 +}}{\text{(aq) + 2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{2Cr}}^{\text{3 +}}{\text{(aq) + 3NpO}}_{\text{2}}^{\text{2 +}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Step 2: Balance all atoms in the equation.

After the addition of the hydronium ions, we need to balance the other atoms as follows, and the final balanced equation is,

${\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2 -}}{\text{(aq) + 3Np}}^{\text{4 +}}{\text{(aq) + 2H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{2Cr}}^{\text{3 +}}{\text{(aq) + 3NpO}}_{\text{2}}^{\text{2 +}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

(d)${\text{HCOOH(aq) + MNO}}_{\text{4}}^{\text{-}}\text{(aq)}\to {\text{CO}}_{\text{2}}{\text{(g) + Mn}}^{\text{2 +}}\text{(aq)}$

Step 1: Balance the oxygen and hydrogen atoms.

In the reactant, there are six oxygen and one hydrogen atom present, but in the product, only two oxygen atoms are present. To balance this atom, we must multiply formic acid by five and manganese oxide by two and add six hydronium ions. The balanced equation is as follows.

${\text{5HCOOH(aq) + 2MNO}}_{\text{4}}^{\text{-}}{\text{(aq) + 6H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{5CO}}_{\text{2}}{\text{(g) + 2Mn}}^{\text{2 +}}{\text{(aq) + 14H}}_{\text{2}}\text{O(l)}$

(e)${\text{Hg}}_{\text{2}}{\text{HPO}}_{\text{4}}{\text{(s) + Au(s) + Cl}}^{\text{-}}\text{(aq)}\to {\text{Hg(l) +H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq) + AuCl}}_{\text{4}}^{\text{-}}\text{(aq)}$

Step 1: Balance hydrogen and chlorine atoms by multiplying threeby HgHPO4 and addingeightchlorine.

To balance the chlorine atoms, we can add eight chlorine atoms into the reactant and multiply three by HgHPO4.

${\text{3Hg}}_{\text{2}}{\text{HPO}}_{\text{4}}{\text{(s) + 2Au(s) + 8Cl}}^{\text{-}}\text{(aq)}\to {\text{6Hg(l) + 3H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq) + 2AuCl}}_{\text{4}}^{\text{-}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Step 2: Balance all atoms.

The balanced equation is as follows

${\text{3Hg}}_{\text{2}}{\text{HPO}}_{\text{4}}{\text{(s) + 2Au(s) + 8Cl}}^{\text{-}}{\text{(aq) + 3H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\to {\text{6Hg(l) + 3H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq) + 2AuCl}}_{\text{4}}^{\text{-}}{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$