Problem 85 - Introductory Chemistry Atoms First

Question

Sodium (Na) reacts with hydrogen \(\left(\mathrm{H}_{2}\right)\) to form sodium hydride (NaH). A reaction mixture contains \(10.00 \mathrm{~g} \mathrm{Na}\) and \(0.0235 \mathrm{~g} \mathrm{H}_{2}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(0.428 \mathrm{~g}\) of \(\mathrm{NaH}\) is recovered. What is the percent yield of the reaction?

Step-by-Step Solution

Answer

(a) The balanced chemical equation for the reaction is: \(2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\) (b) Hydrogen \(\left(\mathrm{H}_{2}\right)\) is the limiting reactant. (c) The theoretical yield for this reaction is 0.5568 g of sodium hydride \(\left(\mathrm{NaH}\right)\). (d) 9.4618 g of excess sodium (Na) is left over at the end of the reaction. (e) The percent yield of the reaction is approximately 76.87%.

Step 1: Write and balance the chemical equation

First, let's write the balanced chemical equation for the reaction between sodium (Na) and hydrogen (H₂) to form sodium hydride (NaH): \[2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\]

Step 2: Determine the limiting reactant

To determine the limiting reactant, we first need to calculate the moles of sodium and hydrogen present in the reaction mixture. Using the molar mass of Na and H₂ (22.99 g/mol and 2.02 g/mol, respectively), we can find the moles of each reactant: Moles of Na = (10.00 g Na) / (22.99 g/mol) = 0.435 mol Na Moles of H₂ = (0.0235 g H₂) / (2.02 g/mol) = 0.0116 mol H₂ Now, divide each by the stoichiometric coefficients from the balanced equation: Moles of Na ratio = (0.435 mol Na) / 2 = 0.218 Moles of H₂ ratio = (0.0116 mol H₂) / 1 = 0.0116 Since 0.0116 is the smallest value, H₂ is the limiting reactant.

Step 3: Calculate the theoretical yield

Using the balanced chemical equation and the moles of the limiting reactant (H₂), we can calculate the theoretical yield of NaH: Moles of NaH = 2 * moles of H₂ = 2 * 0.0116 mol = 0.0232 mol NaH Now, convert moles of NaH to grams using the molar mass of NaH (22.99 + 1.01 = 24 g/mol): Theoretical yield = (0.0232 mol NaH) * (24 g/mol) = 0.5568 g NaH

Step 4: Calculate the amount of excess reactant remaining

To calculate the amount of excess reactant remaining, we need to determine how much Na was consumed. Since 0.0116 mol H₂ was completely consumed, then 0.0232 mol Na (2 * 0.0116 mol) is consumed. Calculate the remaining amount of Na in moles: Moles of Na remaining = 0.435 mol Na (initial) - 0.0232 mol Na (consumed) = 0.4118 mol Na Now, convert the remaining moles of Na into grams using its molar mass (22.99 g/mol): Excess reactant remaining = (0.4118 mol Na) * (22.99 g/mol) = 9.4618 g Na

Step 5: Calculate the percent yield

The percent yield can be calculated by dividing the actual yield (0.428 g NaH) by the theoretical yield (0.5568 g NaH) and multiplying by 100%: Percent yield = (0.428 g NaH / 0.5568 g NaH) * 100% ≈ 76.87%

Key Concepts

Chemical Reactions

In chemical reactions, substances known as reactants are transformed into products. A balanced chemical equation represents these changes and ensures the same number of each type of atom on both sides of the equation. For example, the reaction between sodium (Na) and hydrogen (\(\mathrm{H}_{2}\)) to form sodium hydride (\(\mathrm{NaH}\)) is represented by:
\[2 \mathrm{Na} + \mathrm{H}_{2} \rightarrow 2 \mathrm{NaH}\]
This equation indicates that two atoms of sodium react with one molecule of hydrogen to produce two units of sodium hydride. Balancing the equation is crucial because it reflects the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction.

Limiting Reactant

The concept of a limiting reactant is essential in stoichiometry. It determines how much product can be formed in a chemical reaction. The limiting reactant is the substance that is completely used up first and thus determines the maximum amount of product formed.To find the limiting reactant, we calculate the moles of each reactant based on their masses and molar masses. In our example:

• Moles of Na: \(10.00 \text{ g} / 22.99 \text{ g/mol} = 0.435 \text{ mol Na}\)
• Moles of \(\mathrm{H}_{2}\): \(0.0235 \text{ g} / 2.02 \text{ g/mol} = 0.0116 \text{ mol } \mathrm{H}_{2}\)

To determine which reactant limits the production, we divide these amounts by their coefficients in the balanced equation:

• For Na: \(0.435/2 = 0.218\)
• For \(\mathrm{H}_{2}\): \(0.0116/1 = 0.0116\)

Since \(0.0116\) is smaller, hydrogen is the limiting reactant.

Theoretical Yield

The theoretical yield is the amount of product that would be formed if the reaction went perfectly and completely based on the limiting reactant. It’s an important concept because it represents the maximum possible efficiency of the reaction.Using our balanced equation, we identify the moles of product (NaH) formed from the limiting reactant (\(\mathrm{H}_{2}\)):

• Moles of NaH = \(2 \times 0.0116 = 0.0232\)

To find this in grams, we multiply by the molar mass of NaH (\(24\) g/mol):

• Theoretical Yield = \(0.0232 \text{ mol} \times 24 \text{ g/mol} = 0.5568 \text{ g NaH}\)

This calculation helps in comparing actual results to predicted outcomes to evaluate a reaction's efficiency.

Percent Yield

Percent yield is a measure that compares the actual yield to the theoretical yield, providing insight into the reaction's efficiency. It's calculated using the equation:\[\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\]In our example, the actual yield of NaH was \(0.428 \text{ g}\), and the theoretical yield was \(0.5568 \text{ g}\). Thus, the percent yield is:

• \(\left(\frac{0.428}{0.5568}\right) \times 100\% \approx 76.87\%\)

This measurement is crucial as it accounts for losses and inefficiencies in real-world scenarios, helping chemists understand and improve processes.