Question

Complete and balance the following equations. If no reaction occurs, write NR

(a) $H_{3}P{O}_4\left(l\right)+NaI\left(s\right)\to$

(b) $C{l}_2\left(g\right)+I^{-}\left(aq\right)\to$

(c) $B{r}_2\left(l\right)+C{l}^{-}\left(aq\right)\to$

(d) $ClF\left(g\right)+F_2\left(g\right)\to$

Short answer

(a) $H_{3}P{O}_4\left(l\right)+NaI\left(s\right)\to$

(b) $C{l}_2\left(g\right)+2I^{-}\left(aq\right)\to I_2\left(g\right)+2C{l}^{-}\left(aq\right)$

(c) $B{r}_2+C{l}^{-}\to NR$

(d) $ClF\left(g\right)+F_2\left(g\right)\to Cl{F}_3\left(g\right)$

Step-by-step solution

Chemical reactions

Processes that involve the formation of a substance (product) from another substance (reactant) involving the breaking and forming of chemical bonds is called a chemical reaction. The chemical identity of the reactant is changed during a chemical reaction.

The occurrence of a chemical reaction depends on the reactivity of a substance and the energy of the reactant and the products.

Balanced chemical equations

A chemical equation represents a chemical reaction and is considered balanced if it obeys the law of conservation of mass. Therefore, no mass is created or destroyed during a chemical reaction, and the number of atoms of all the elements will be equal on the reactant and product sides.

A chemical equation is balanced by the trial method in which coefficients are added to the reactants and products, such that the number of atoms on the two sides becomes equal. The coefficients of reactants and products in a balanced equation are stoichiometric values.

Balancing the given equation

(a)

Phosphoric acid reacts with sodium iodide to form hydrogen iodide and sodium phosphate. It is an example of a double displacement reaction, and the chemical equation can be shown as

$H_{3}P{O}_4+NaI\to HI+N{a}_{3}P{O}_4$

The number of atoms on the reactant side and product side can be shown as

Element	Reactant side	Product side
H	3	1
P	1	1
O	4	4
Na	1	3
I	1	1

Hydrogen and sodium are not balanced in the equation. It can be balanced by adding a coefficient of 3 to hydrogen iodide and sodium iodide.

Therefore, the balanced equation for the reaction is

$H_{3}P{O}_4\left(l\right)+3NaI\left(aq\right)\to 3HI\left(aq\right)+N{a}_{3}P{O}_4\left(aq\right)$

(b)

Chlorine can displace iodide ions from its solution, causing iodine gas evolution. Chlorine moves into the solution. The chemical equation can be shown as

$C{l}_2+I^{-}\to I_2+C{l}^{-}$

The number of atoms on each side can be shown as

Element	Reactant side	Product side
Cl	2	1
I	1	2

Coefficients are added to equate the number of atoms on each side. Therefore, the balanced equation for the reaction can be shown as

$C{l}_2\left(g\right)+2I^{-}\left(aq\right)\to I_2\left(g\right)+2C{l}^{-}\left(aq\right)$

(c)

The reactivity of halogens decreases down the group. A more reactive element can displace a less reactive element from the solution. Therefore, a given halogen can displace only the halogens below it in the periodic table.

Therefore, bromine can displace iodide ions from the solution, but not chloride ions or fluoride ions. Then, the addition of bromine to chloride solution will have no effect, and no reaction takes place.

$B{r}_2+C{l}^{-}\to NR$

(d)

Chlorine monofluoride reacts with fluorine gas to form chlorine trifluoride. It is an addition reaction given by the chemical equation:

$ClF+F_2\to Cl{F}_3$

The number of atoms of each element can be shown as

Element	Reactant side	Product side
Cl	1	1
F	3	3

The number of atoms of all elements is equal on both sides. Therefore, the equation is already balanced.

Therefore, the balanced equation for the reaction is

$ClF\left(g\right)+F_2\left(g\right)\to Cl{F}_3\left(g\right)$