Question

Complete and balance the following equations: (a) \(\mathrm{NaOCH}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\) (b) \(\mathrm{CuO}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow\) (c) \(\mathrm{WO}_{3}(s)+\mathrm{H}_{2}(g) \stackrel{\Delta}{\longrightarrow}\) (d) \(\mathrm{NH}_{2} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \longrightarrow\) (e) \(\mathrm{Al}_{4} \mathrm{C}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow\)

Short answer

(a) NaOCH3(s) + H2O(l) → NaOH(aq) + CH3OH(l) (b) CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l) (c) WO3(s) + 3 H2(g) → W(s) + 3 H2O(l) (d) 4 NH2OH(l) + 3 O2(g) → 2 N2(g) + 6 H2O(l) (e) Al4C3(s) + 12 H2O(l) → 4 Al(OH)3(s) + 3 CH4(g)

Step-by-step solution

(a) Balancing NaOCH3(s) + H2O(l) → ?

First, we need to complete the equation by identifying the products formed. In this reaction, sodium methoxide (NaOCH3) acts as a base, and water (H2O) acts as an acid. Sodium hydroxide (NaOH) and methanol (CH3OH) are formed as products. Now, we can balance the equation: NaOCH3(s) + H2O(l) → NaOH(aq) + CH3OH(l) The equation is already balanced with equal numbers of atoms for each element on both sides.

(b) Balancing CuO(s) + HNO3(aq) → ?

Copper(II) oxide (CuO) reacts with nitric acid (HNO3) in an acid-base reaction, forming copper(II) nitrate (Cu(NO3)2) and water (H2O) as products. Now, we can balance the equation: CuO(s) + 2 HNO3(aq) → Cu(NO3)2(aq) + H2O(l)

(c) Balancing WO3(s) + H2(g) → ?

Tungsten(VI) oxide (WO3) reacts with hydrogen gas (H2) in a reduction reaction. The products are tungsten metal (W) and water (H2O). Now, we can balance the equation by using a heat symbol (∆) to denote that heat is applied to the reaction: WO3(s) + 3 H2(g) → W(s) + 3 H2O(l)

(d) Balancing NH2OH(l) + O2(g) → ?

Hydroxylamine (NH2OH) reacts with oxygen gas (O2) in a combustion reaction. The products are nitrogen gas (N2) and water (H2O). Balanced combustion reactions typically have whole-number coefficients. Now, we can balance the equation: 2 NH2OH(l) + 3/2 O2(g) → N2(g) + 3 H2O(l) To make all coefficients whole numbers, we can multiply the whole equation by 2: 4 NH2OH(l) + 3 O2(g) → 2 N2(g) + 6 H2O(l)

(e) Balancing Al4C3(s) + H2O(l) → ?

Aluminum carbide (Al4C3) reacts with water (H2O) to form aluminum hydroxide (Al(OH)3) and methane gas (CH4) as products. Now, we can balance the equation: Al4C3(s) + 12 H2O(l) → 4 Al(OH)3(s) + 3 CH4(g)

Key concepts

Balancing Reactions

Balancing chemical reactions is essential to ensure that the same number of each type of atom appears on both sides of the equation. This adherence to the law of conservation of mass allows chemists to know the exact proportions of reactants and products involved in a reaction.

To balance a chemical equation, follow these steps:

• Write down the unbalanced equation.
• List the number of atoms for each element present in the reactants and the products.
• Adjust the coefficients (the numbers in front of each substance) to balance the atoms of each element.
• Recheck to ensure the same number of atoms for each element is present on both sides of the equation.

For reactions involving complex molecules or requiring trial and error, start with elements that appear in the least compounds, and leave elements like oxygen and hydrogen for last. This systematic approach simplifies balancing reactions, making it easier to get accurate results.

Acid-Base Reactions

Acid-base reactions are a fundamental type of chemical reaction that involves the transfer of protons ( ext{H}^+ ext{ ions ext}). These reactions occur when an acid donates a proton to a base.

Some classic examples include reactions involving acids like ext{HNO}_3 and bases such as ext{NaOH}. In the given exercise:

• ext{CuO}(s) + 2 ext{HNO}_3(aq) → ext{Cu(NO}_3 ext{)}_2(aq) + ext{H}_2 ext{O}(l): Copper(II) oxide acts as a base, while nitric acid acts as the acid.

Balancing these reactions involves ensuring the charges and the number of atoms are equal on both sides. It can require adjusting coefficients, especially when dealing with polyatomic ions as units.

One practical tip is to balance the compounds' primary atoms first and adjust the ions' coefficients, which often results in the correct number of ext{H}^+ ext{ and } ext{OH}^- ext{ ions ext} to form water.

Reduction Reactions

Reduction reactions involve the gain of electrons by a molecule, atom, or ion. This process is usually paired with oxidation, where another substance loses electrons—completing what is known as a redox reaction.

In these reactions, you will often see a change in the oxidation state of the substances involved. For instance:

• When ext{WO}_3(s) + 3 ext{H}_2(g) → ext{W}(s) + 3 ext{H}_2 ext{O}(l), tungsten(VI) oxide is reduced to tungsten metal as it gains electrons from hydrogen gas.

Identifying the substances that undergo either reduction or oxidation is key. You can do this by following these steps:

• Assign oxidation numbers to each atom in the reactants and products.
• Identify which atoms' oxidation numbers increase or decrease.
• The element with the decreased oxidation number is reduced, while the one with increased oxidation number is oxidized.

This understanding helps in balancing redox reactions by ensuring electron transfer is accounted for.

Combustion Reactions

Combustion reactions involve an organic compound reacting with oxygen gas ( ext{O}_2) resulting in the production of carbon dioxide ( ext{CO}_2), water ( ext{H}_2 ext{O}), and often energy in the form of heat and light.

In the exercise example:

• ext{4 NH}_2 ext{OH}(l) + 3 ext{O}_2(g) → 2 ext{N}_2(g) + 6 ext{H}_2 ext{O}(l), the compound hydroxylamine combusts (or oxidizes) with oxygen.

To balance combustion reactions, aim to balance carbon, hydrogen, and then oxygen atoms, in that order. Adjust the coefficients of the reactants and products until each element has the same number on both sides of the equation.

Often, combustion reactions in organic chemistry involve long chains of hydrocarbons, so balancing might require iterations to get the correct stoichiometric coefficients. Don't forget to convert fractions into whole numbers for coefficients, ensuring they reflect the simplest whole-number ratio.