Problem 14 - Chemistry The Central Science In Si Units

Question

Balance the following equations: (a) \(\mathrm{CF}_{4}(l)+\mathrm{Br}_{2}(g) \longrightarrow \mathrm{CBr}_{4}(l)+\mathrm{F}_{2}(g)\) (b) \(\mathrm{Cu}(s)+\mathrm{HNO}_{3}(a q) \longrightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (c) \(\mathrm{MnO}_{2}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{MnCl}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Cl}_{2}(g)\) (d) \(\mathrm{KOH}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \longrightarrow \mathrm{K}_{3} \mathrm{PO}_{4}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

(a) \(2\mathrm{CF}_{4} + 4\mathrm{Br}_{2} \rightarrow \mathrm{CBr}_{4} + 2\mathrm{F}_{2}\); (b) \(\mathrm{Cu} + 4\mathrm{HNO}_{3} \rightarrow \mathrm{Cu(NO}_{3})_{2} + 2\mathrm{NO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\); (c) \(\mathrm{MnO}_{2} + 4\mathrm{HCl} \rightarrow \mathrm{MnCl}_{2} + \mathrm{H}_{2}\mathrm{O} + \mathrm{Cl}_{2}\); (d) \(3\mathrm{KOH} + \mathrm{H}_{3}\mathrm{PO}_{4} \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4} + 3\mathrm{H}_{2}\mathrm{O}\).

Step 1: Write the Unbalanced Equation (a)

The equation is given as \(\mathrm{CF_{4}}(l) + \mathrm{Br_{2}}(g) \rightarrow \mathrm{CBr_{4}}(l) + \mathrm{F_{2}}(g)\). This is the starting point for the balancing process.

Step 2: Balance for Carbon and CBr (a)

Since there is 1 carbon atom on both sides in \(\mathrm{CF_{4}}\) and \(\mathrm{CBr_{4}}\), carbon is already balanced.

Step 3: Balance for Bromine and Fluorine (a)

The left side has 2 Br in \(\mathrm{Br_{2}}\), but the product \(\mathrm{CBr_{4}}\) contains 4 Br atoms. Balance by adjusting coefficients: \(\mathrm{2\ CF_{4}}(l) + \mathrm{4\ Br_{2}}(g) \rightarrow \mathrm{CBr_{4}}(l) + \mathrm{2\ F_{2}}(g)\). Now, each element has equal counts on both sides.

Step 4: Write the Unbalanced Equation (b)

The equation is \(\mathrm{Cu}(s) + \mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Cu(NO}_{3})_2(aq) + \mathrm{NO}_{2}(g) + \mathrm{H}_{2}\mathrm{O}(l)\). We start here for balancing.

Step 5: Balance Copper and Nitrate Ions (b)

Place a 3 before \(\mathrm{HNO}_{3}\) to balance nitrate ions: \(\mathrm{Cu}(s) + 4\mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Cu(NO}_{3})_2(aq) + \mathrm{NO}_2(g) + \mathrm{H}_{2}\mathrm{O}(l)\).

Step 6: Balance for Hydrogen and Oxygen (b)

Balance hydrogen and oxygen by adjusting water: With 2 oxygen from 2 water molecules and 4 from nitrogens on both sides, complete the balance: \(\mathrm{Cu}(s) + 4\mathrm{HNO}_{3}(aq) \rightarrow \mathrm{Cu(NO}_{3})_2(aq) + 2\mathrm{NO}_{2}(g) + 2\mathrm{H}_{2}\mathrm{O}(l)\).

Step 7: Write the Unbalanced Equation (c)

The given chemical equation is \(\mathrm{MnO}_{2}(s) + \mathrm{HCl}(aq) \rightarrow \mathrm{MnCl}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Cl}_{2}(g)\).

Step 8: Balance Manganese (c)

Manganese atom count is 1 on both sides in \(\mathrm{MnO}_{2}\) and \(\mathrm{MnCl}_{2}\). Thus, they are balanced.

Step 9: Balance for Chlorine and Oxygen (c)

There are 2 chlorine atoms in \(\mathrm{Cl}_2\) and \(\mathrm{MnCl}_2\), multiply \(\mathrm{HCl}\) by 4: \(\mathrm{MnO}_{2}(s) + 4\mathrm{HCl}(aq) \rightarrow \mathrm{MnCl}_{2}(s) + \mathrm{H}_{2}\mathrm{O}(l) + \mathrm{Cl}_{2}(g)\). Oxygen from \(\mathrm{MnO}_{2}\) is balanced with water.

Step 10: Write the Unbalanced Equation (d)

The given equation is \(\mathrm{KOH}(aq) + \mathrm{H}_{3}\mathrm{PO}_{4}(aq) \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4}(aq) + \mathrm{H}_{2}\mathrm{O}(l)\).

Step 11: Balance Potassium and Phosphate (d)

Balance potassium and phosphate by using a coefficient 3 before \(\mathrm{KOH}\): \(3\mathrm{KOH}(aq) + \mathrm{H}_{3}\mathrm{PO}_{4}(aq) \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4}(aq) + \mathrm{H}_{2}\mathrm{O}(l)\).

Step 12: Confirm Balance for Hydrogen and Oxygen (d)

Place a coefficient 3 in front of \(\mathrm{H}_{2}\mathrm{O}\) to balance hydrogen, thereby aligning oxygen with phosphate: the final balanced equation is \(3\mathrm{KOH}(aq) + \mathrm{H}_{3}\mathrm{PO}_{4}(aq) \rightarrow \mathrm{K}_{3}\mathrm{PO}_{4}(aq) + 3\mathrm{H}_{2}\mathrm{O}(l)\).

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative aspects of chemical reactions. It involves calculating the relative quantities of reactants and products involved in a reaction. This is essential when balancing chemical equations because it ensures that the proportions of each element are correctly adjusted to comply with the law of conservation of mass.

• The first step is to interpret the given chemical equation and identify the number of atoms or molecules for each element involved.
• Stoichiometry relies on using coefficients in front of chemical formulas to signify the number of moles of a substance.
• Accurate stoichiometric calculations allow chemists to predict the amounts of substances consumed and formed.

By ensuring the stoichiometry is correct, you are guaranteeing the conservation of mass in the reaction, as the amount of each element remains constant from reactants to products, just in different forms.

Chemical Reactions

Chemical reactions are processes where reactants undergo a transformation to form products. They are characterized by the breaking and forming of chemical bonds. To better understand chemical reactions, one must consider the nature of the substances involved and how they interact under certain conditions.

• Reactions are represented by chemical equations that show the reactants turning into products.
• In a balanced chemical reaction, the mass and number of atoms are conserved.
• Various types of chemical reactions include synthesis, decomposition, single replacement, and double replacement reactions.
• Identifying the type of reaction can guide how to balance the equation and predict products.

Understanding these reactions can help in predicting how different substances can be transformed or combined in specific conditions.

Conservation of Mass

At the heart of chemical reactions lies the law of conservation of mass. This principle states that mass is neither created nor destroyed in a chemical reaction. This means that the mass of the reactants must equal the mass of the products.

• This fundamental law ensures that all chemical reactions must be balanced.
• The total mass of all atoms present in the reaction must stay constant before and after the reaction.
• This is why balancing equations is crucial: to reflect that atoms are merely rearranged.

By adhering to the conservation of mass, you confirm that chemical equations accurately depict the transformation of substances. It means you can predict the amounts required and produced, leading to meaningful observations and discoveries in chemical processes.