Problem 12 - Chemistry and Chemical Reactivity

Question

Balance the following equations, and name each reactant and product: (a) \(\mathrm{SF}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{HF}(\ell)\) (b) \(\mathrm{NH}_{3}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)\) (c) \(\mathrm{BF}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq})\)

Step-by-Step Solution

Answer

(a) SF4(g) + 2H2O(l) → SO2(g) + 4HF(l); (b) 4NH3(aq) + 5O2(aq) → 4NO(g) + 6H2O(l); (c) BF3(g) + 3H2O(l) → 3HF(aq) + H3BO3(aq).

Step 1: Identify reactants and products for equation (a)

For the equation \( \mathrm{SF}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{HF}(\ell) \), the reactants are sulfur tetrafluoride (\( \mathrm{SF}_4 \)) and water (\( \mathrm{H}_2\mathrm{O} \)). The products are sulfur dioxide (\( \mathrm{SO}_2 \)) and hydrogen fluoride (\( \mathrm{HF} \)).

Step 2: Balance equation (a)

Start by balancing sulfur. There is one \( \mathrm{S} \) atom on both sides, so sulfur is balanced. Next, balance oxygen. There is one \( \mathrm{O} \) atom on the reactant side and two on the product side, so place a coefficient of 2 in front of \( \mathrm{H}_2\mathrm{O} \):\[ \mathrm{SF}_{4}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{HF}(\ell) \]Next, balance hydrogen. There are 4 hydrogen atoms in the reactants and only 2 in the product, so place a coefficient of 4 in front of \( \mathrm{HF} \):\[ \mathrm{SF}_{4}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g}) + 4\mathrm{HF}(\ell) \]Finally, check fluorine. There are 4 fluorine atoms in \( \mathrm{SF}_4 \) and 4 in \( 4\mathrm{HF} \), so fluorine is balanced. The balanced equation is:\[ \mathrm{SF}_{4}(\mathrm{g}) + 2\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{SO}_{2}(\mathrm{g}) + 4\mathrm{HF}(\ell) \]

Step 3: Identify reactants and products for equation (b)

For \( \mathrm{NH}_{3}(\mathrm{aq})+\mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{NO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \), the reactants are ammonia (\( \mathrm{NH}_3 \)) and oxygen (\( \mathrm{O}_2 \)), and the products are nitric oxide (\( \mathrm{NO} \)) and water (\( \mathrm{H}_2\mathrm{O} \)).

Step 4: Balance equation (b)

Start with nitrogen. There is one nitrogen atom on both sides, so nitrogen is balanced. Balance oxygen next. There are 2 oxygen atoms in \( \mathrm{O}_2 \) and one in \( \mathrm{NO} \) and \( \mathrm{H}_2\mathrm{O} \), which adds up to 3. To balance, put a coefficient of 2 in front of \( \mathrm{NO} \) and \( \mathrm{H}_2\mathrm{O} \):\[ \mathrm{NH}_{3}(\mathrm{aq}) + \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \]Now balance nitrogen again. There's 1 nitrogen in the reactant and 2 in the products, so put a coefficient of 2 in front of \( \mathrm{NH}_3 \):\[ 2 \mathrm{NH}_{3}(\mathrm{aq}) + \mathrm{O}_{2}(\mathrm{aq}) \rightarrow 2 \mathrm{NO}(\mathrm{g}) + 2 \mathrm{H}_{2} \mathrm{O}(\ell) \] Finally, balance hydrogen. There are 6 hydrogen atoms in \( 2 \mathrm{NH}_3 \) and 4 in \( 2 \mathrm{H}_2 \mathrm{O} \). Add an extra H by changing coefficients to:\[ 4 \mathrm{NH}_{3}(\mathrm{aq}) + 5\mathrm{O}_{2}(\mathrm{aq}) \rightarrow 4\mathrm{NO}(\mathrm{g}) + 6 \mathrm{H}_{2} \mathrm{O}(\ell) \]Finally, there are 4 oxygen from NO and 6 from water, so this configuration balances.

Step 5: Identify reactants and products for equation (c)

In the equation \( \mathrm{BF}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HF}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq}) \), the reactants are boron trifluoride (\( \mathrm{BF}_3 \)) and water (\( \mathrm{H}_2\mathrm{O} \)), while the products are hydrofluoric acid (\( \mathrm{HF} \)) and boric acid (\( \mathrm{H}_3\mathrm{BO}_3 \)).

Step 6: Balance equation (c)

Start by balancing boron. There is one \( \mathrm{B} \) atom on both sides, so boron is balanced. Next, balance fluorine. There are 3 fluorine atoms in \( \mathrm{BF}_3 \), so place a coefficient of 3 in front of \( \mathrm{HF} \):\[ \mathrm{BF}_{3}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 3 \mathrm{HF}(\mathrm{aq}) + \mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq}) \]Lastly, balance hydrogen. There are now 3 hydrogens in \( \mathrm{H}_3\mathrm{BO}_3 \) and 3 in \( 3 \mathrm{HF} \), totaling 6 on the product side. Therefore, put a coefficient of 3 in front of \( \mathrm{H}_2\mathrm{O} \) to match:\[ \mathrm{BF}_{3}(\mathrm{g}) + 3 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 3 \mathrm{HF}(\mathrm{aq}) + \mathrm{H}_{3} \mathrm{BO}_{3}(\mathrm{aq}) \]All elements are now balanced.

Key Concepts

Chemical Reactants and Products

In any chemical reaction, the substances that undergo a chemical change are known as reactants. The substances that are produced as a result of this change are called products. Returning to our provided equations:

• In equation (a), the reactants are sulfur tetrafluoride (\(\mathrm{SF}_4\)) and water (\(\mathrm{H}_2\mathrm{O}\)). These react with each other to form the products, sulfur dioxide (\(\mathrm{SO}_2\)) and hydrogen fluoride (\(\mathrm{HF}\)).
• For equation (b), the reactants are ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)), yielding products nitric oxide (\(\mathrm{NO}\)) and water (\(\mathrm{H}_2\mathrm{O}\)).
• In equation (c), boron trifluoride (\(\mathrm{BF}_3\)) and water are the reactants, with hydrofluoric acid (\(\mathrm{HF}\)) and boric acid (\(\mathrm{H}_3\mathrm{BO}_3\)) as the products.

Understanding these terms is vital as it helps in tracking elements through a reaction, giving insight into how reactants transform into new products.

Stoichiometry

Stoichiometry is the study of the quantitative aspects of chemical formulas and reactions. It is essentially the calculation of reactants and products in a chemical equation, which is crucial for ensuring equations are balanced correctly. In simpler terms, stoichiometry tells us the proportions in which substances react and produce new substances.
By using stoichiometry, we can predict:

• The amount of products that will form in a chemical reaction
• The amount of reactants needed to produce a desired amount of product
• The leftovers after a reaction, if any

This concept is grounded on the principle of the conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Thus, the total mass of reactants equals the mass of products in any given reaction.
Stoichiometry allows us to translate these quantities into measurable amounts, typically in moles, and provides the foundation for balancing chemical reactions.

Chemical Equations Balancing Steps

Balancing a chemical equation involves making sure that the number of each type of atom on the reactant side is equal to the number on the product side. This is essential as it reflects the conservation of mass.
Here’s a simplified step-by-step approach to balance chemical equations:

• Identify all reactants and products: Start by writing down the unbalanced equation. Clearly label each side as reactants and products.
• List the number of atoms of each element: Count the number of atoms of each element in both reactants and products.
• Balance one element at a time: Use coefficients to multiply the number of molecules so that each type of atom has the same number on both sides of the equation. Start with elements that appear in only one reactant and one product first.
• Double-check each element: Verify that all elements are balanced, adjusting the coefficients as necessary. Make sure the coefficients are in the simplest whole number ratio.
• Do a final check: Ensure that the number of atoms for each element is the same on both sides to validate the equation is balanced.

Mastering these steps requires practice but is pivotal for a thorough understanding of chemical reactions and processes.