Question

Balance these equations. (a) Reaction of calcium cyanamide to produce ammonia: $$\mathrm{CaNCN}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\ell) \longrightarrow \mathrm{CaCO}_{3}(\mathrm{~s})+\mathrm{NH}_{3}(\mathrm{~g}) $$ (b) Reaction to produce diborane, \(\mathrm{B}_{2} \mathrm{H}_{6}:\) $$ \mathrm{NaBH}_{4}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq}) \longrightarrow \mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})+\mathrm{Na}_{2} \mathrm{SO}_{4}(\mathrm{aq}) $$ (c) Reaction to rid water of hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) a foulsmelling compound: $$ \mathrm{H}_{2} \mathrm{~S}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow \mathrm{S}_{8}(\mathrm{~s})+\mathrm{HCl}(\mathrm{aq}) $$

Short answer

(a) \( \mathrm{CaNCN} + 3\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2\mathrm{NH}_{3} \). (b) \( 2\mathrm{NaBH}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{B}_{2} \mathrm{H}_{6} + 4\mathrm{H}_{2} + \mathrm{Na}_{2} \mathrm{SO}_{4} \). (c) \( 8\mathrm{H}_{2} \mathrm{S} + 16\mathrm{Cl}_{2} \rightarrow \mathrm{S}_{8} + 16\mathrm{HCl} \).

Step-by-step solution

Balance Equation (a) - Identify Elements and Count Atoms

Start by identifying the elements in the equation \( \mathrm{CaNCN} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + \mathrm{NH}_{3} \). The elements present are: Calcium (Ca), Nitrogen (N), Carbon (C), Hydrogen (H), and Oxygen (O). Count the atoms on both sides: - Left: Ca = 1, N = 2, C = 1, H = 2, O = 1- Right: Ca = 1, C = 1, N = 1, H = 3, O = 3.

Balance Equation (a) - Adjust Nitrogen Atoms

To balance the nitrogen atoms, notice there are 2 nitrogen atoms on the left and only 1 on the right. Add a coefficient of 2 in front of \( \mathrm{NH}_{3} \) on the right side, changing the NH3 count to two nitrogen atoms and six hydrogen atoms:\[ \mathrm{CaNCN} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2\mathrm{NH}_{3} \]

Balance Equation (a) - Adjust Hydrogen Atoms

Now, balance the hydrogen atoms. You have 2 hydrogen in water on the left and 6 in the ammonia on the right. You'll need 3 water molecules to provide 6 hydrogen atoms:\[ \mathrm{CaNCN} + 3\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2\mathrm{NH}_{3} \]

Balance Equation (a) - Verify All Elements

Make sure that all elements are balanced:- Ca: 1 on both sides- N: 2 on both sides- C: 1 on both sides- H: 6 on both sides- O: 3 on both sidesThe balanced equation is:\[ \mathrm{CaNCN} + 3\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2\mathrm{NH}_{3} \]

Balance Equation (b) - Identify Elements and Count Atoms

For the reaction \( \mathrm{NaBH}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{B}_{2} \mathrm{H}_{6} + \mathrm{H}_{2} + \mathrm{Na}_{2} \mathrm{SO}_{4} \), the elements are Na, B, H, S, and O. Count the atoms:- Left: Na = 1, B = 1, H = 6, S = 1, O = 4- Right: Na = 2, B = 2, H = 8, S = 1, O = 4.

Balance Equation (b) - Adjust Sodium and Boron

To balance sodium (Na) and boron (B), place a coefficient of 2 in front of \( \mathrm{NaBH}_{4} \) on the left:\[ 2\mathrm{NaBH}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{B}_{2} \mathrm{H}_{6} + \mathrm{H}_{2} + \mathrm{Na}_{2} \mathrm{SO}_{4} \]

Balance Equation (b) - Adjust Hydrogen Atoms

There are now 10 hydrogen atoms (8 from \( \mathrm{B}_{2} \mathrm{H}_{6} \) and \( \mathrm{H}_{2} \), and 2 from \( \mathrm{NaBH}_{4} \)). Balance by adjusting the \( \mathrm{H}_{2} \) coefficient to 4:\[ 2\mathrm{NaBH}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{B}_{2} \mathrm{H}_{6} + 4\mathrm{H}_{2} + \mathrm{Na}_{2} \mathrm{SO}_{4} \]

Balance Equation (b) - Verify

Ensure all elements are balanced:- Na: 2 on both sides- B: 2 on both sides- H: 14 on both sides- S: 1 on both sides- O: 4 on both sidesFinal balanced reaction is:\[ 2\mathrm{NaBH}_{4} + \mathrm{H}_{2} \mathrm{SO}_{4} \rightarrow \mathrm{B}_{2} \mathrm{H}_{6} + 4\mathrm{H}_{2} + \mathrm{Na}_{2} \mathrm{SO}_{4} \]

Balance Equation (c) - Identify Elements and Count Atoms

For \( \mathrm{H}_{2} \mathrm{S} + \mathrm{Cl}_{2} \rightarrow \mathrm{S}_{8} + \mathrm{HCl} \), count the atoms:- Left: S = 1, H = 2, Cl = 2- Right: S = 8, H = 1, Cl = 1.

Balance Equation (c) - Adjust Sulfur Atoms

Add a coefficient of 8 in front of \( \mathrm{H}_{2} \mathrm{S} \) on the left to match the 8 sulfur atoms on the right:\[ 8\mathrm{H}_{2} \mathrm{S} + \mathrm{Cl}_{2} \rightarrow \mathrm{S}_{8} + \mathrm{HCl} \]

Balance Equation (c) - Adjust Hydrogen and Chlorine Atoms

To balance hydrogen and chlorine, place a 16 in front of \( \mathrm{HCl} \) on the right and a 16 in front of \( \mathrm{Cl}_{2} \) on the left:\[ 8\mathrm{H}_{2} \mathrm{S} + 16\mathrm{Cl}_{2} \rightarrow \mathrm{S}_{8} + 16\mathrm{HCl} \]

Balance Equation (c) - Verify

Check to ensure all elements are balanced:- Sulfur (S): 8 on both sides- Hydrogen (H): 16 on both sides- Chlorine (Cl): 32 on both sidesThe balanced reaction is:\[ 8\mathrm{H}_{2} \mathrm{S} + 16\mathrm{Cl}_{2} \rightarrow \mathrm{S}_{8} + 16\mathrm{HCl} \]

Key concepts

Stoichiometry

Stoichiometry is the heart of chemical balancing. It involves calculating the quantities of reactants and products in a chemical reaction. Understand stoichiometry, and you'll master chemical equations.
When balancing equations, stoichiometry ensures that the law of conservation of mass is satisfied. This means the mass and number of atoms of each element are equal on both sides of the equation.
Consider the equation for ammonia production: \( \mathrm{CaNCN} + \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + \mathrm{NH}_{3} \).
Using stoichiometry, here are steps to balance:

• Identify elements: Ca, N, C, H, O.
• Count atoms for each element on both sides.
• Adjust coefficients to balance the atoms across the equation.

By repeating these steps, you achieve a balanced equation, which allows for clear, calculable reactions.
Exercising such stoichiometric calculations helps in predicting the amount of products formed from given reactants or vice-versa.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances, called products.
The key to understanding reactions lies in understanding how molecules in reactants break and re-form in products.
In the reactions given:

• (a) Ammonia is formed from calcium cyanamide reacting with water.
• (b) Diborane is produced by reacting sodium borohydride with sulfuric acid.
• (c) Sulfur precipitates from the reaction of hydrogen sulfide with chlorine.

Each equation follows the law of conservation of mass where atoms are neither created nor destroyed.
Recognizing reaction types is crucial, as it tells us more about the nature and energy changes happening within the reaction.
For example, type (c) is a redox (reduction-oxidation) reaction where chlorine gains electrons (is reduced), forming hydrochloric acid and elemental sulfur.

Inorganic Chemistry

Inorganic chemistry focuses on compounds that are not carbon-based, such as metals, salts, and minerals.
In each of the given reactions, inorganic compounds are the main players:

• Calcium cyanamide and water are inorganic, forming inorganic products like calcium carbonate.
• Sodium borohydride, used in reaction (b), is a common inorganic reducing agent.
• Reaction (c) neutralizes hydrogen sulfide—a nonmetal gas—using chlorine, forming hydrochloric acid and elemental sulfur.

In this subfield, understanding how elements and simple compounds interact is key.
Balancing these reactions assists chemists in predicting how different inorganic substances react and what compounds they might produce.
It paves the way for applications like fertilizer production in reaction (a), hydrogen storage and fuel in reaction (b), and gas purification in reaction (c).
The heart of inorganic chemistry lies in such versatile reactions and their balanced chemical equations.