Problem 13 - Chemistry Matter and Change

Question

Calculate Solve each of the following. \begin{equation} \begin{array}{l}{\text { a. In the complete reaction of } 22.99 \mathrm{g} \text { of sodium with } 35.45 \mathrm{g} \text { of chlorine, what }} \\\ {\text { mass of sodium chloride is formed? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { b. } A 12.2 \text { -g sample of } X \text { reacts with a sample of } Y \text { to form } 78.9 \text { g of } X Y . \text { What }} \\ {\text { is the mass of } Y \text { that reacted? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. 58.44 g of NaCl is formed. b. 66.7 g of Y reacted.

Step 1: Identify the Reaction for Part (a)

For part (a), we are dealing with the reaction between sodium (Na) and chlorine (Cl) to form sodium chloride (NaCl). The balanced chemical equation is: \[2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl}\]

Step 2: Calculate the Moles of Reactants for Part (a)

Find the moles of sodium and chlorine using their respective molar masses: - Molar mass of sodium (Na) = 22.99 g/mol- Moles of sodium = \(\frac{22.99 \text{ g}}{22.99 \text{ g/mol}} = 1 \text{ mol}\)- Molar mass of chlorine (Cl) = 35.45 g/mol, but chlorine is diatomic (Cl₂), so it should be 70.90 g/mol- Moles of chlorine = \(\frac{35.45 \text{ g}}{70.90 \text{ g/mol}} = 0.5 \text{ mol} \text{ Cl}_2\)

Step 3: Determine Limiting Reactant for Part (a)

According to the balanced equation, 2 moles of Na react with 1 mole of Cl₂ to form 2 moles of NaCl. Since we have 1 mole of Na and 0.5 mole of Cl₂, neither is in excess for 1 mole of Na; all reactants will fully combine to form NaCl.

Step 4: Calculate Mass of Sodium Chloride for Part (a)

Since all reactants are used up completely, moles of NaCl formed is equal to moles of Na used, which is 1 mole. Sodium chloride (NaCl), has a molar mass of 58.44 g/mol. Mass of NaCl = moles of NaCl × molar mass of NaCl = \(1 \text{ mol} \times 58.44 \text{ g/mol} = 58.44 \text{ g}\)

Step 5: Summarize Calculation for Part (a)

The complete reaction of 22.99 g of sodium with 35.45 g of chlorine results in the formation of 58.44 g of NaCl.

Step 6: Set Up Equation for Part (b)

For part (b), we have a reaction where element X reacts with element Y to form a compound XY, resulting in a product with a mass of 78.9 g. We know the mass of X is 12.2 g.

Step 7: Calculate Mass of Reactant Y for Part (b)

Assuming conservation of mass in the reaction, the mass of reactants (X + Y) equals the mass of the products (XY). Therefore, the equation for mass is:\[ \text{mass of } X + \text{mass of } Y = \text{mass of } XY \]Substituting the known values:\[ 12.2 \text{ g} + \text{mass of } Y = 78.9 \text{ g} \]Solving for the mass of Y gives:\[ \text{mass of } Y = 78.9 \text{ g} - 12.2 \text{ g} = 66.7 \text{ g} \]

Step 8: Conclusion for Part (b)

The mass of element Y that reacted in forming compound XY is 66.7 g.

Key Concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines the amount of product formed. It is the reactant that is used up completely during the reaction. Understanding the limiting reactant is essential because it limits the amount of product that can be produced.
In our example, sodium (Na) and chlorine (Cl) react to form sodium chloride (NaCl). Looking at the stoichiometry of the balanced equation, \[2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl},\]we see that two moles of sodium react with one mole of chlorine. In this case, we initially have one mole of sodium and 0.5 moles of chlorine.
Since two moles of sodium are required to react completely with one mole of chlorine, and we have precisely the amounts needed based on their stoichiometric ratios, both reactants are consumed entirely, indicating neither excess nor limiting reactant. This simple ratio check ensures the reaction will go to completion using all available materials.

Molar Mass Calculation

Molar mass is a key term in stoichiometry referring to the mass of one mole of a given substance. It's critical for converting between grams and moles. To calculate the molar mass of a compound or an element, sum the atomic masses of all the constituent atoms. Atomic masses can be found on the periodic table.
In our exercise, we needed the molar masses for sodium (Na), chlorine (Cl, diatomic as Cl₂), and sodium chloride (NaCl).

• Sodium (Na) has a molar mass of 22.99 g/mol.
• Chlorine (Cl₂) is diatomic, so its molar mass is 2 times 35.45 g/mol, which equals 70.90 g/mol.
• Sodium chloride (NaCl) has a molar mass given by summing the molar masses of Na and Cl, equaling 58.44 g/mol.

This calculation is foundational in shifting between mass and mole measurements, enabling accurate predictions of product yield in reactions.

Mass Conservation

The principle of mass conservation states that in a closed system, mass cannot be created or destroyed, only rearranged. This principle is fundamental to stoichiometry and chemical reactions. It dictates that the total mass of reactants must equal the total mass of products.
In our exercise's second part, this concept was used to find the mass of element Y that reacted. By knowing the total mass of product XY (78.9 g) and the known mass of element X (12.2 g), we can conclude, using equations:
\[\text{mass of } X + \text{mass of } Y = \text{mass of } XY \]Substituting:\[12.2 \text{ g} + \text{mass of } Y = 78.9 \text{ g}\]We determined the mass of Y to be 66.7 g. Such problems underscore how mass conservation helps us account for substances when not all reactant quantities are directly provided.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show how reactants are transformed into products, using chemical formulas and stoichiometric coefficients that represent the relative amounts of each substance involved.
The balanced equation for sodium and chlorine reacting is \[2 \text{Na} + \text{Cl}_2 \rightarrow 2 \text{NaCl}.\]Balance is vital, as it obeys the mass conservation principle, ensuring that atoms are neither lost nor gained in the reaction. For each element, the number of atoms on the reactant side should be equal to the number on the product side.

• "2 Na" means two sodium atoms are involved.
• "Cl₂" means the chlorine is diatomic, existing naturally as pairs of atoms.
• The "2 NaCl" indicates each side of the equation has two sodium and two chlorine atoms, aligning with stoichiometric rules.

Writing and balancing chemical equations is fundamental in predicting and understanding chemical reactions.