Question

3.58 Write balanced equations for each of the following by inserting the correct coefficients in the blanks: (a) \(-\mathrm{Cu}(s)+\underline{\mathrm{S}}_{8}(s) \longrightarrow-\mathrm{Cu}_{2} \mathrm{~S}(s)\) (b) \(-\mathrm{P}_{4} \mathrm{O}_{10}(s)+\underline{\mathrm{H}}_{2} \mathrm{O}(l) \longrightarrow-\mathrm{H}_{3} \mathrm{PO}_{4}(l)\) (c) \(-\mathrm{B}_{2} \mathrm{O}_{3}(s)+-\mathrm{NaOH}(a q) \longrightarrow-\mathrm{Na}_{3} \mathrm{BO}_{3}(a q)+-\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(-\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\underline{\mathrm{O}_{2}}(g) \longrightarrow\) \(-\mathrm{CO}_{2}(g)+-\mathrm{H}_{2} \mathrm{O}(g)+-\mathrm{N}_{2}(g)\)

Short answer

(a) 8Cu + \text{S}_8 \rightarrow 4\text{Cu}_2\text{S} (b) \text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4 (c) \text{B}_2\text{O}_3 + 6\text{NaOH} \rightarrow 2\text{Na}_3\text{BO}_3 + 3\text{H}_2\text{O} (d) 4\text{CH}_3\text{NH}_2 + 9\text{O}_2 \rightarrow 4\text{CO}_2 + 10\text{H}_2\text{O} + 2\text{N}_2

Step-by-step solution

- Balance the equation for part (a)

Start by balancing the sulfur atoms. In \(\text{S}_8\), there are 8 sulfur atoms. Thus, on the product side, we need 8 sulfur atoms, which means we need 4 \(\text{Cu}_2\text{S}\) because there is 1 sulfur in each \(\text{Cu}_2\text{S}\). Now balance the copper atoms. On the product side, there are \(4 \times 2 = 8\), so you need 8 \(\text{Cu}\) atoms on the reactant side. The balanced equation is: \[8\text{Cu}(s) + \text{S}_8(s) \rightarrow 4\text{Cu}_2\text{S}(s)\]

- Balance the equation for part (b)

First balance the phosphorus atoms. There are 4 phosphorus atoms in \(\text{P}_4\text{O}_{10}\), so we need 4 \(\text{H}_3\text{PO}_4\). This gives us \(4 \times 3 = 12\) hydrogen atoms. Balance the hydrogen atoms by placing 6 in front of \(\text{H}_2\text{O}\). The balanced equation is: \[ \text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4\text{H}_3\text{PO}_4(l) \]

- Balance the equation for part (c)

Start by balancing the boron atoms. There are 2 borons in \(\text{B}_2\text{O}_3\), so we need 2 \(\text{Na}_3\text{BO}_3\). Balance the sodium atoms by placing 6 in front of \(\text{NaOH}\). Finally, balance the oxygen and hydrogen atoms. You will need 3 \(\text{H}_2\text{O}\) to balance the leftover hydrogen atoms. The balanced equation is: \[ \text{B}_2\text{O}_3(s) + 6\text{NaOH}(aq) \rightarrow 2\text{Na}_3\text{BO}_3(aq) + 3\text{H}_2\text{O}(l) \]

- Balance the equation for part (d)

Start by balancing the carbon atoms. There is 1 carbon atom in \(\text{CH}_3\text{NH}_2\), so we need 1 \(\text{CO}_2 \). Next, balance the nitrogen atoms. There is 1 nitrogen in \(\text{N}_2\) and 1 in \(\text{CH}_3\text{NH}_2\), so put 2 in front of the nitrogen gas to balance it. Balance the hydrogen and oxygen atoms by placing 2 in front of \(\text{H}_2\text{O}\) and 3.5 in front of \(\text{O}_2\). The balanced equation is: \[ 4\text{CH}_3\text{NH}_2(g) + 9\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 10\text{H}_2\text{O}(g) + 2\text{N}_2(g) \]

Key concepts

Law of Conservation of Mass

The Law of Conservation of Mass states that in a closed system, mass is conserved during a chemical reaction. This means that the mass of the reactants equals the mass of the products. In essence, matter cannot be created or destroyed.

When we're balancing chemical equations, this law is our foundational guide. We balance the equations so that the same number of each type of atom is present on both sides of the equation.

For example, in part (a) of our exercise, we balanced the equation:
8Cu(s) + S8(s) → 4Cu2S(s). By ensuring each side has eight copper atoms and eight sulfur atoms, we adhered to the Law of Conservation of Mass.

Every balanced equation respects this law, making it a key principle in chemistry.

Chemical Reactions

Chemical reactions involve rearranging atoms to form new substances. Reactants are the starting substances, and products are the substances formed.

In our exercise, we have different types of chemical reactions. For part (b) - P4O10(s) + 6H2O(l) → 4H3PO4(l) - it's a synthesis reaction where simpler substances combine to form a more complex product.

Similarly, part (c) - B2O3(s) + 6NaOH(aq) → 2Na3BO3(aq) + 3H2O(l) - is a type of double displacement reaction. Two compounds exchange components to form new compounds.

Understanding these reactions helps us predict the products and balance the equations accurately.

Stoichiometry

Stoichiometry is the quantitative aspect of chemical reactions. It involves calculations based on the balanced equations, ensuring we adhere to the Law of Conservation of Mass.

By using molar ratios derived from the coefficients in a balanced equation, we can determine the amount of reactants needed or products formed.

Take part (d) from our exercise: 4CH3NH2(g) + 9O2(g) → 4CO2(g) + 10H2O(g) + 2N2(g). Here, stoichiometry tells us that four moles of methane amine react with nine moles of oxygen to produce four moles of carbon dioxide, ten moles of water, and two moles of nitrogen gas.

Balancing these equations allows us to perform accurate stoichiometric calculations in chemical reactions.

Reactants and Products

Reactants are substances consumed during a chemical reaction, while products are the resulting substances.

In our exercise, each part showcases different reactants and products:

• (a) Reactants: Cu and S8, Products: Cu2S
• (b) Reactants: P4O10 and H2O, Products: H3PO4
• (c) Reactants: B2O3 and NaOH, Products: Na3BO3 and H2O
• (d) Reactants: CH3NH2 and O2, Products: CO2, H2O, and N2

By understanding the nature of these reactants and products, we can predict how they interact and ensure we balance them correctly in chemical equations.

This foundational knowledge aids us in solving complex problems and performing precise laboratory experiments.