Problem 116 - Chemistry and Chemical Reactivity

Question

Sodium borohydride, NaBH_, reduces many metal ions to the metal. (a) Write a balanced equation for the reaction of \(\mathrm{NaBH}_{4}\) with \(\mathrm{AgNO}_{3}\) in water to give silver metal, \(\mathrm{H}_{2}\) gas, boric acid, and sodium nitrate. (The chemistry of \(\mathrm{NaBH}_{4}\) is described in Section 21.6.) (b) What mass of silver can be produced from \(575 \mathrm{mL}\) of \(0.011 \mathrm{M} \mathrm{AgNO}_{3}\) and \(13.0 \mathrm{g}\) of \(\mathrm{NaBH}_{4} ?\)

Step-by-Step Solution

Answer

0.6823 grams of silver can be produced.

Step 1: Write Unbalanced Chemical Equation

Identify the reactants and products. The reactants are sodium borohydride \(\mathrm{NaBH}_{4}\) and silver nitrate \(\mathrm{AgNO}_{3}\). The products are silver metal \(\mathrm{Ag}\), hydrogen gas \(\mathrm{H}_{2}\), boric acid \(\mathrm{H}_{3} \mathrm{BO}_{3}\), and sodium nitrate \(\mathrm{NaNO}_{3}\). Write the unbalanced equation:\[\mathrm{NaBH}_{4} + \mathrm{AgNO}_{3} \rightarrow \mathrm{Ag} + \mathrm{H}_{2} + \mathrm{H}_{3}\mathrm{BO}_{3} + \mathrm{NaNO}_{3}\]

Step 2: Balance the Chemical Equation

Balance each element one at a time by adjusting coefficients. Begin with Ag ions, followed by Na, B, H, and O, since these appear in multiple compounds:1. Balance Ag by placing a coefficient 1 before \(\mathrm{Ag}\).2. Balance Na: One \(\mathrm{Na}\) on both sides.3. Balance B: One \(\mathrm{B}\) on both sides.4. Balance H: Four H from \(\mathrm{NaBH}_{4}\) match with \(1\, \mathrm{H}_2\) and 3 from \(\mathrm{H}_{3} \mathrm{BO}_{3}\).5. Balance O and NO from the nitrate groups.The balanced equation is:\[\mathrm{4NaBH}_{4} + 16\mathrm{AgNO}_{3} + 6\mathrm{H}_{2} \rightarrow 16\mathrm{Ag} + 4\mathrm{NaNO}_{3} + 4\mathrm{H}_{3}\mathrm{BO}_{3} + 22\mathrm{H}_{2}\]

Step 3: Calculate Moles of Reactants

Using molarities and volumes to find moles of \( \mathrm{AgNO}_{3} \):- The volume of \( \mathrm{AgNO}_{3} \) is given as 575 mL, or 0.575 L.- Molarity is 0.011 M, so moles of \( \mathrm{AgNO}_{3} \) = \(0.575 \times 0.011 = 0.006325 \text{ moles}\).Calculate moles of \( \mathrm{NaBH}_{4} \):- Molar mass of \( \mathrm{NaBH}_{4} \) is approximately 37.83 g/mol.- Mass given is 13.0 g, so moles of \( \mathrm{NaBH}_{4} \) = \( \frac{13.0}{37.83} \approx 0.3439 \text{ moles}\).

Step 4: Determine Limiting Reactant

From the balanced equation, 1 mole of \( \mathrm{NaBH}_{4} \) reacts with 4 moles of \( \mathrm{AgNO}_{3} \).- Required moles of \( \mathrm{AgNO}_{3} \) for 0.3439 moles of \( \mathrm{NaBH}_{4} \) is \( 0.3439 \times 4 = 1.3756 \text{ moles}\).- Since \( \mathrm{AgNO}_{3} \) available is only 0.006325 moles, \( \mathrm{AgNO}_{3} \) is the limiting reactant.

Step 5: Calculate Mass of Silver Produced

From the balanced equation, 1 mole of \( \mathrm{AgNO}_{3} \) produces 1 mole of \( \mathrm{Ag} \).- Moles of silver \( \mathrm{Ag} \) = moles of \( \mathrm{AgNO}_{3} \) = 0.006325 moles.- Molar mass of silver (\( \mathrm{Ag} \)) is approximately 107.87 g/mol.- Mass of silver = \( 0.006325 \times 107.87 \approx 0.6823 \text{ grams}\).

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry. It's like ensuring that a recipe has the correct proportions of ingredients. Balancing ensures that the number of atoms for each element is the same on both sides of the equation. This stems from the law of conservation of mass which states that matter cannot be created or destroyed.

Here's how you do it:

• Write the unbalanced equation first, noting all reactants and products.
• Adjust the coefficients (numbers in front of the compounds) to ensure the same number of each type of atom appears on both sides.
• Start by balancing the most complex molecule.
• Leave hydrogen and oxygen for last as they often appear in multiple compounds.

By balancing the chemical equation for the reaction between sodium borohydride and silver nitrate, we ensure the reaction reflects reality and obeys the conservation laws.

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that gets used up first, stopping the reaction. Identifying the limiting reactant ensures you know how much product can be formed.

Here's how you figure it out:

• Calculate the moles of each reactant available.
• Use the stoichiometric coefficients from the balanced equation to determine which reactant produces the least amount of product.
• The reactant that produces the least amount of product is the limiting reactant.

In our example, after calculating the available moles, we found that silver nitrate (\( \mathrm{AgNO}_{3} \)) is the limiting reactant, as we require more of it than is available to react with the sodium borohydride present.

Stoichiometry

Stoichiometry involves using balanced chemical equations to calculate the relationships between the reactants and products. It provides the quantitative "recipe" that tells how much of each component is needed or produced.

Steps to use stoichiometry:

• Use the balanced chemical equation to determine the molar relationships between reactants and products.
• Calculate the moles of substances available or required.
• Apply conversion factors based on the balanced equation to find masses or volumes of interest.

In our exercise, we used stoichiometry to calculate the moles of silver that could be produced, based on the limiting reactant available, translating that into a mass measurement.

Reduction Reactions

Reduction reactions involve the gaining of electrons by a substance. In the context of our exercise, sodium borohydride reduces silver ions to metallic silver. This means silver ions gain electrons and are converted into their elemental metal form.

Key aspects of reduction reactions:

• Reduction is the gain of electrons, while oxidation is the loss.
• In a redox reaction, one substance is reduced and another is oxidized.
• The element that gains electrons undergoes reduction, symbolizing a decrease in oxidation state.

In our example, the silver ions (\( \mathrm{Ag}^+ \)) gain electrons to form silver metal (\( \mathrm{Ag} \)), showcasing a classic reduction.