Question

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

Short answer

\( \begin{aligned} \text{(a)} \,\, &\mathrm{NaOCH}_3(\mathrm{s}) + \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \longrightarrow \mathrm{NaOH}(\mathrm{aq}) + \mathrm{CH}_3 \mathrm{OH}(\mathrm{l}) \\ \text{(b)} \,\, &\mathrm{CuO}(\mathrm{s}) + 2\mathrm{HNO}_3(\mathrm{aq}) \longrightarrow \mathrm{Cu(NO}_3)_2(\mathrm{aq}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) \\ \text{(c)} \,\, &\mathrm{WO}_3(\mathrm{s}) + 3\mathrm{H}_2(\mathrm{g}) \stackrel{\Delta}{\longrightarrow} \mathrm{W}(\mathrm{s}) + 3\mathrm{H}_2\mathrm{O}(\mathrm{l}) \\ \text{(d)} \,\, &4\mathrm{NH}_2\mathrm{OH}(\mathrm{l}) + 3\mathrm{O}_2(\mathrm{g}) \longrightarrow 4\mathrm{H}_2\mathrm{O}(\mathrm{l}) + 2\mathrm{N}_2(\mathrm{g}) \\ \text{(e)} \,\, &\mathrm{Al}_4\mathrm{C}_3(\mathrm{s}) + 12\mathrm{H}_2\mathrm{O}(\mathrm{l}) \longrightarrow 4\mathrm{Al(OH)}_3(\mathrm{s}) + 3\mathrm{CH}_4(\mathrm{g}) \end{aligned} \)

Step-by-step solution

Identify the products

Sodium methoxide (NaOCH3) reacts with water (H2O) to form sodium hydroxide (NaOH) and methanol (CH3OH).

Balance the equation

The balanced equation would be: NaOCH3(s) + H2O(l) -> NaOH(aq) + CH3OH(l) The equation is already balanced, with one Na, one O, three H, and one C on both sides. #(b) Completing and Balancing the Reaction of Copper(II) Oxide and Nitric Acid#

Identify the products

Copper(II) oxide (CuO) reacts with nitric acid (HNO3) to form copper(II) nitrate (Cu(NO3)2) and water (H2O).

Balance the equation

The balanced equation would be: CuO(s) + 2HNO3(aq) -> Cu(NO3)2(aq) + H2O(l) Here, one Cu, one O, two N, and four H atoms are present on both sides. #(c) Completing and Balancing the Reaction of Tungsten(VI) Oxide and Hydrogen#

Identify the products

Tungsten(VI) oxide (WO3) reacts with hydrogen gas (H2) under heating to form tungsten (W) and water (H2O).

Balance the equation

The balanced equation would be: WO3(s) + 3H2(g) -> W(s) + 3H2O(l) In this equation, one W, three O, and six H atoms are present on both sides. #(d) Completing and Balancing the Reaction of Hydroxylamine and Oxygen#

Identify the products

Hydroxylamine (NH2OH) reacts with oxygen gas (O2) to form water (H2O) and nitrogen gas (N2).

Balance the equation

The balanced equation would be: 4NH2OH(l) + 3O2(g) -> 4H2O(l) + 2N2(g) Here, four N, eight H, and six O atoms are present on both sides. #(e) Completing and Balancing the Reaction of Aluminum Carbide and Water#

Identify the products

Aluminum carbide (Al4C3) reacts with water (H2O) to form aluminum hydroxide (Al(OH)3) and methane (CH4).

Balance the equation

The balanced equation would be: Al4C3(s) + 12H2O(l) -> 4Al(OH)3(s) + 3CH4(g) In this equation, four Al, twelve O, four C, and thirty-six H atoms are present on both sides.

Key concepts

Sodium Methoxide Reaction

Sodium methoxide is a compound that reacts easily with water in a chemical reaction. During this reaction, each molecule of sodium methoxide (\(\text{NaOCH}_3\)) interacts with a molecule of water (\(\text{H}_2\text{O}\)). This interaction results in the formation of sodium hydroxide (\(\text{NaOH}\)) and methanol (\(\text{CH}_3\text{OH}\)). Here’s a simple breakdown:

• Sodium methoxide provides the sodium ion and methoxide ion.
• Water supplies the hydrogen and hydroxyl ions.

The balanced chemical equation for this reaction is \(\text{NaOCH}_3(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{NaOH}(\text{aq}) + \text{CH}_3\text{OH}(\text{l})\).
Notice that each of the atoms of sodium, oxygen, hydrogen, and carbon is balanced on both sides of the equation. This balance ensures that the law of conservation of mass is upheld. Understanding and balancing such equations is crucial for predicting the outcomes of chemical reactions.

Copper(II) Oxide Reaction

Copper(II) oxide is a black solid that reacts with nitric acid, a common laboratory acid. When these two reactants come together, they form copper(II) nitrate, a blue-green compound, and water. This is an example of a chemical reaction that involves a metal oxide and an acid, leading to a salt and water:

• Copper(II) oxide contributes copper ions and oxide ions.
• Nitric acid provides protons and nitrate ions.

The balanced equation for this reaction is \(\text{CuO}(\text{s}) + 2\text{HNO}_3(aq) \rightarrow \text{Cu(NO}_3\text{)}_2(aq) + \text{H}_2\text{O}(\text{l})\).
The equation is balanced with respect to all elements—ensuring each type of atom is equal on both sides reflects the conservation of mass. This reaction is useful in generating copper nitrates, which are important in chemistry.

Tungsten(VI) Oxide Reaction

Tungsten(VI) oxide is a heavy compound that can be reduced to tungsten metal by reacting with hydrogen gas under high temperatures. This type of reaction is termed a reduction, as the hydrogen helps to pull away the oxygen atoms from the tungsten oxide:

• Tungsten(VI) oxide starts with a tungsten ion bound to three oxygen atoms.
• Hydrogen gas supplies protons that combine with oxygen to form water.

The reaction’s balanced equation is \(\text{WO}_3(\text{s}) + 3\text{H}_2(\text{g}) \overset{\Delta}{\rightarrow} \text{W}(\text{s}) + 3\text{H}_2\text{O}(\text{l})\).
This balanced representation shows that for every molecule of tungsten(VI) oxide, three molecules of hydrogen gas are used to produce one atom of tungsten metal and three molecules of water. Learning to balance such equations is key in predicting the quantities of products formed.

Hydroxylamine Reaction

Hydroxylamine is a versatile compound that reacts with oxygen gas to yield nitrogen gas and water. This reaction exemplifies a redox reaction:

• Hydroxylamine donates hydrogen atoms.
• Oxygen gas receives these hydrogen atoms to form water.

The balanced equation is \(4\text{NH}_2\text{OH}(\text{l}) + 3\text{O}_2(\text{g}) \rightarrow 4\text{H}_2\text{O}(\text{l}) + 2\text{N}_2(\text{g})\).
In this equation, it is essential to recognize the changes in oxidation states and appropriately distribute the molecules on both sides to ensure balance. This results in a conservation of nitrogen and oxygen atoms. Mastery of redox reactions is fundamental in understanding chemical energy and bonds.

Aluminum Carbide Reaction

Aluminum carbide reacts with water in an interesting chemical reaction that produces aluminum hydroxide and methane gas. This reaction highlights the interaction between a metal carbide and water:

• Aluminum carbide consists of aluminum and carbon atoms.
• Water molecules undergo reactions to supply hydroxide ions and hydrogen atoms.

The balanced equation for this process is \(\text{Al}_4\text{C}_3(\text{s}) + 12\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{Al(OH)}_3(\text{s}) + 3\text{CH}_4(\text{g})\).
This reaction’s equation balances thirty-six hydrogen atoms across both sides, among others, ensuring all atoms match, thus conserving mass. Understanding the balancing in such reactions is pivotal for both chemical synthesis and predictive calculations in laboratory or industrial settings.