Problem 137 - Chemistry

Question

The following reaction is stoichiometric as written$$\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}+\mathrm{NaOC}_{2}\mathrm{H}_{5} \longrightarrow \mathrm{C}_{4}\mathrm{H}_{8}+\mathrm{C}_{2} \mathrm{H}_{5}\mathrm{OH}+\mathrm{NaCl}$$ but it is often carried out with an excess of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) to react with any water present in the reaction mixture that might reduce the yield. If the reaction shown was carried out with \(6.83 \mathrm{~g}\) of \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\), how many grams of \(\mathrm{NaOC}_{2} \mathrm{H}_{5}\) would be needed to have a 50 percent molar excess of that reactant?

Step-by-Step Solution

Answer

A 50 percent molar excess of \(NaOC_2H_5\) equates to an approximate mass of 7.79 g

Step 1: Calculate Moles of Given Reactant

The molar mass of \(C_4H_9Cl\) is calculated by adding up the molar masses of its constituent atoms. This results in a molar mass of approximately 108.6 g/mol. Therefore, the amount in moles of \(C_4H_9Cl\) that we have can be calculated using its mass and its molar mass: Moles of \(C_4H_9Cl = \frac{Mass}{Molar Mass} = \frac{6.83 g}{108.6 g/mol} = 0.063 mol

Step 2: Calculate Moles of Required Reactant

From the balanced chemical equation, we can see that the molar ratio of \(NaOC_2H_5\) to \(C_4H_9Cl\) is 1:1. However, we are asked to calculate the amount of \(NaOC_2H_5\) required for a 50% molar excess. Therefore, the moles of \(NaOC_2H_5\) required would be 1.5 times the moles of \(C_4H_9Cl\). Hence, Moles of \(NaOC_2H_5\) required = 1.5 * Moles of \(C_4H_9Cl\) = 1.5 * 0.063 mol = 0.095 mol

Step 3: Calculate Mass of Required Reactant

Once we have the moles of \(NaOC_2H_5\) required, we can then convert it into mass using its molar mass. The molar mass of \(NaOC_2H_5\) is approximately 82 g/mol. Therefore, the mass of \(NaOC_2H_5\) required will be Moles * Molar Mass = 0.095 mol * 82 g/mol = 7.79 g

Key Concepts

Molar Mass

Understanding molar mass is crucial in stoichiometry. It's the mass of one mole of a substance, typically measured in grams per mole (g/mol). To find the molar mass of a compound, sum up the molar masses of all the atoms present in the formula. These molar masses are taken from the periodic table.
For example, in the compound \(C_4H_9Cl\), calculate as follows:

• Carbon (C): with a molar mass of 12.01 g/mol, and there are 4 atoms of carbon in \(C_4H_9Cl\), contributing \(4 \times 12.01 =48.04\) g/mol.
• Hydrogen (H): with a molar mass of 1.01 g/mol, and 9 hydrogens are present, contributing \(9 \times 1.01 =9.09 \) g/mol.
• Chlorine (Cl): with a molar mass of 35.45 g/mol, contributing \(35.45\) g/mol.

This results in a total molar mass of approximately 108.6 g/mol. Knowing this allows us to relate masses to moles - an essential aspect of stoichiometry.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. These processes follow the principles of conservation of mass and energy. In a balanced chemical equation, the number of atoms for each element are identical on both the reactant and product sides.
In our example reaction:

• The reactants \(\, C_4H_9Cl + NaOC_2H_5\), turn into the products \(\, C_4H_8 + C_2H_5OH + NaCl\).
• Each reactant and product are formed in a 1:1 molar ratio.

Balancing means ensuring these ratios are upheld; no atom is lost or gained, just rearranged. This balance is vital for accurately calculating reactant and product quantities, as seen when we convert moles of \(NaOC_2H_5\) to grams in the previous solution.

Limiting Reactant

The concept of a limiting reactant is key in determining the amount of product formed in a reaction. This reactant is completely consumed in the reaction, limiting the extent and amount of product that can be created.
In some reactions, like the one provided, it's common to include excess of one reactant to ensure another, more valuable or sensitive reactant isn't the limiting factor. Here, \(NaOC_2H_5\) is used in a 50% molar excess.
Excess reactant:

• Allows compensation for other reactants lost due to impurities, side reactions, or moisture.
• Ensures the main reactant, here \(C_4H_9Cl\), fully reacts to produce the desired outcome.

Understanding and calculating limiting reactants and using excess wisely becomes crucial in industrial and laboratory settings to maximize yield and efficiency.