Problem 105 - Chemistry: The Central Science in SI Units, Global Edition

Question

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resulting combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Step-by-Step Solution

Answer

(a) The balanced chemical equation for the reaction is: \[2\mathrm{C}_{2}\mathrm{H}_{2} + 5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] (b) The limiting reactant is oxygen \(\left(\mathrm{O}_{2}\right)\). (c) After the reaction is complete, the amounts in grams of each substance are: \(\mathrm{C}_{2}\mathrm{H}_{2}\): \(6.74\, \mathrm{g}\), \(\mathrm{O}_{2}\): \(0\, \mathrm{g}\), \(\mathrm{CO}_{2}\): \(11.0\, \mathrm{g}\), and \(\mathrm{H}_{2}\mathrm{O}\): \(2.25\, \mathrm{g}\).

Step 1: (a) Write the balanced chemical equation

The combustion reaction of acetylene \(\left(\mathrm{C}_{2}\mathrm{H}_{2}\right)\) with oxygen \(\left(\mathrm{O}_{2}\right)\) produces carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) and water \(\left(\mathrm{H}_{2}\mathrm{O}\right)\). First, write an unbalanced chemical equation. \[\mathrm{C}_{2}\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\] Next, balance the chemical equation. There are two carbons, two hydrogens, and two oxygens on the left side and one carbon, two hydrogens, and three oxygens on the right side. The balanced chemical equation is: \[\mathrm{C}_{2}\mathrm{H}_{2} + 2.5\mathrm{O}_{2} \rightarrow 2\mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\] However, fractions in stoichiometry are unusual, so it's preferable to use whole numbers. To achieve this, multiply the entire reaction by 2: \[2\mathrm{C}_{2}\mathrm{H}_{2} + 5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\] This balanced chemical equation represents the reaction.

Step 2: (b) Find the limiting reactant

To determine the limiting reactant, we need to know the amount of each reactant in moles. Use the molar mass to convert grams to moles. The molar mass of \(\mathrm{C}_{2}\mathrm{H}_{2}\) is 26.04 g/mol, and that of \(\mathrm{O}_{2}\) is 32.00 g/mol. For \(\mathrm{C}_{2}\mathrm{H}_{2}\): \[10.0 \mathrm{~g}\, \mathrm{C}_{2}\mathrm{H}_{2} \times \frac{1\, \text{mol} \, \mathrm{C}_{2}\mathrm{H}_{2}}{26.04\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}} \approx 0.384\, \mathrm{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] For \(\mathrm{O}_{2}\): \[10.0 \mathrm{~g}\, \mathrm{O}_{2} \times \frac{1\, \text{mol}\, \mathrm{O}_{2}}{32.00\, \mathrm{g}\, \mathrm{O}_{2}} \approx 0.312\, \text{mol}\, \mathrm{O}_{2}\] Now, calculate the mole ratios by dividing the moles of each substance by its stoichiometric coefficient: For \(\mathrm{C}_{2}\mathrm{H}_{2}\): \(\frac{0.384}{2} = 0.192\) For \(\mathrm{O}_{2}\): \(\frac{0.312}{5} = 0.062\) The smallest mole ratio value corresponds to the limiting reactant, which is \(\mathrm{O}_{2}\).

Step 3: (c) Calculate amounts after the reaction is complete

Since oxygen is the limiting reactant, we can calculate the amount of acetylene remaining and the amounts of carbon dioxide and water formed using the balanced equation. 1. Calculate moles of acetylene remaining: From the balanced equation, 2 moles of \(\mathrm{C}_{2}\mathrm{H}_{2}\) react with 5 moles of \(\mathrm{O}_{2}\). Therefore, \(0.312\, \text{mol}\, \mathrm{O}_{2}\) will react with: \[0.312\, \text{mol}\, \mathrm{O}_{2} \times \frac{2\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.125\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] Subtract the reacted \(\mathrm{C}_{2}\mathrm{H}_{2}\) from the initial amount: \[0.384\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} - 0.125\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} \approx 0.259\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2}\] Convert moles to grams: \[0.259\, \text{mol}\, \mathrm{C}_{2}\mathrm{H}_{2} \times 26.04\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}\, \text{mol}^{-1} \approx 6.74\, \mathrm{g}\, \mathrm{C}_{2}\mathrm{H}_{2}\] 2. Calculate moles and grams of carbon dioxide produced: From the balanced equation, 4 moles of \(\mathrm{CO}_{2}\) are produced for every 5 moles of \(\mathrm{O}_{2}\) reacted: \[0.312\, \text{mol}\, \mathrm{O}_{2} \times \frac{4\, \text{mol}\, \mathrm{CO}_{2}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.250\, \text{mol}\, \mathrm{CO}_{2}\] Convert moles to grams: \[0.250\, \text{mol}\, \mathrm{CO}_{2} \times 44.01\, \mathrm{g}\, \mathrm{CO}_{2}\, \text{mol}^{-1} \approx 11.0\, \mathrm{g}\, \mathrm{CO}_{2}\] 3. Calculate moles and grams of water produced: From the balanced equation, 2 moles of \(\mathrm{H}_{2}\mathrm{O}\) are produced for every 5 moles of \(\mathrm{O}_{2}\) reacted: \[0.312\, \mathrm{mol}\, \mathrm{O}_{2} \times \frac{2\, \text{mol}\, \mathrm{H}_{2}\mathrm{O}}{5\, \text{mol}\, \mathrm{O}_{2}} \approx 0.125\, \text{mol}\, \mathrm{H}_{2}\mathrm{O}\] Convert moles to grams: \[0.125\, \text{mol}\, \mathrm{H}_{2}\mathrm{O} \times 18.02\, \mathrm{g}\, \mathrm{H}_{2}\mathrm{O}\, \text{mol}^{-1} \approx 2.25\, \mathrm{g}\, \mathrm{H}_{2}\mathrm{O}\] Final amounts after the reaction is complete: \(\mathrm{C}_{2}\mathrm{H}_{2}\): \(6.74\, \mathrm{g}\) \(\mathrm{O}_{2}\): \(0\, \mathrm{g}\) (limiting reactant, all reacted) \(\mathrm{CO}_{2}\): \(11.0\, \mathrm{g}\) \(\mathrm{H}_{2}\mathrm{O}\): \(2.25\, \mathrm{g}\)

Key Concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial for understanding why reactions stop before all reactants are consumed. It is the reactant that is completely used up first, stopping the reaction because no further product can be formed without it.
To determine the limiting reactant, you must consider the stoichiometry of the reaction. This involves:

• Calculating the number of moles of each reactant available.
• Comparing the mole ratios from the balanced equation to find which reactant runs out first.

In our example, 10.0 grams of oxygen (\(\mathrm{O}_{2}\)) and 10.0 grams of acetylene (\(\mathrm{C}_{2}\mathrm{H}_{2}\)) are used. By converting these masses to moles and comparing their ratios according to the balanced chemical equation, we identified oxygen as the limiting reactant. This information allows us to predict the amounts of products formed and how much of the other reactant remains after the reaction.

Balanced Chemical Equation

A balanced chemical equation is fundamental in chemistry, ensuring that the law of conservation of mass is obeyed. This principle states that matter cannot be created or destroyed, so the mass of reactants must equal the mass of products.
Balancing equations involves:

• Ensuring the same type and number of atoms are present on each side of the equation.
• Adjusting coefficients in front of chemical formulas to achieve balance.

For the combustion of acetylene with oxygen, the unbalanced equation is:\[\mathrm{C}_{2}\mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O}\]When balanced, it reads:\[2\mathrm{C}_{2}\mathrm{H}_{2} + 5\mathrm{O}_{2} \rightarrow 4\mathrm{CO}_{2} + 2\mathrm{H}_{2}\mathrm{O}\]Here, multiplying the entire equation by two ensures no fractions appear, optimizing the stoichiometric calculations.

Stoichiometry

Stoichiometry is the method used in chemistry to calculate the quantities of reactants and products in a chemical reaction. It's all about the balanced equation and understanding the proportional relationships between different substances involved.
Using stoichiometry involves:

• Determining molar masses of reactants and products.
• Converting given masses into moles to use in calculations.
• Applying mole ratios from the balanced equation to predict quantities of products and left-over reactants.

In our exercise, we started with known masses of \(\mathrm{C}_{2}\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\). We used stoichiometry to find out how much of each reactant would react and the exact amounts of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\) produced, providing a comprehensive understanding of the entire reaction process.