Problem 105 - Chemistry The Central Science In Si Units

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) \(2\mathrm{C}_2\mathrm{H}_2 + 5\mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 2\mathrm{H}_2\mathrm{O}\), (b) \(\mathrm{O}_2\) is the limiting reactant, (c) \(6.76\,\mathrm{g}\,\mathrm{C}_2\mathrm{H}_2\), \(0\,\mathrm{g}\,\mathrm{O}_2\), \(11.0\,\mathrm{g}\,\mathrm{CO}_2\), and \(4.5\,\mathrm{g}\,\mathrm{H}_2\mathrm{O}\).

Step 1: Write the Balanced Chemical Equation

First, determine the balanced chemical equation for the combustion of acetylene (\( \mathrm{C}_2\mathrm{H}_2 \)) in oxygen (\( \mathrm{O}_{2} \)). The general form for combustion of a hydrocarbon is \( \mathrm{C}_x\mathrm{H}_y + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \). For acetylene: \[ 2 \mathrm{C}_2\mathrm{H}_2 + 5 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O} \]. This is the balanced equation.

Step 2: Calculate the Moles of Each Reactant

Using the molar masses: \( \mathrm{C}_2\mathrm{H}_2 = 26 \mathrm{~g/mol} \) and \( \mathrm{O}_2 = 32 \mathrm{~g/mol} \), calculate moles for each reactant. For acetylene: \( \frac{10.0 \mathrm{~g}}{26 \mathrm{~g/mol}} \approx 0.385 \mathrm{~mol} \). For oxygen: \( \frac{10.0 \mathrm{~g}}{32 \mathrm{~g/mol}} \approx 0.313 \mathrm{~mol} \).

Step 3: Identify the Limiting Reactant

According to the balanced equation, the molar ratio required is 2:5 for \( \mathrm{C}_2\mathrm{H}_2 : \mathrm{O}_2 \). We have \(0.385 \) mol \( \mathrm{C}_2\mathrm{H}_2 \) and \(0.313 \) mol \( \mathrm{O}_2 \). Calculate required \( \mathrm{O}_2 \) for \( \mathrm{C}_2\mathrm{H}_2\): \( 5 \times 0.385 / 2 = 0.963 \mathrm{~mol} \), which exceeds available \( \mathrm{O}_2 \). Thus, \( \mathrm{O}_2 \) is limiting.

Step 4: Calculate Mass of Reactants and Products After Reaction

Given \( \mathrm{O}_2 \) is limiting, use its moles to find product masses. Reacting \(0.313 \mathrm{~mol} \mathrm{O}_2\) consumes \(0.125 \mathrm{~mol} \mathrm{C}_2\mathrm{H}_2\) (from stoichiometry 5:2), leaving \(0.385 - 0.125 = 0.260 \mathrm{~mol} \mathrm{C}_2\mathrm{H}_2\). \(0.125 \mathrm{~mol} \mathrm{C}_2\mathrm{H}_2\) yields \( 2 \times 0.125 = 0.25 \mathrm{~mol} \mathrm{CO}_2\) (from 2:4 stoichiometry). \( \mathrm{CO}_2 \) mass: \(0.25 \times 44 \mathrm{~g/mol} = 11.0 \mathrm{~g}\). Water: \(\mathrm{H}_2\mathrm{O} = 0.125 \times 2 = 0.25 \mathrm{~mol}\) (from 2:2 stoichiometry), mass: \(0.25 \times 18 \mathrm{~g/mol} = 4.5 \mathrm{~g}\). Remaining acetylene: \(0.26 \times 26 \mathrm{~g/mol} = 6.76 \mathrm{~g}\). Oxygen is completely consumed.

Key Concepts

Combustion Reaction

In a combustion reaction, a substance reacts with oxygen to release heat, forming new products, often carbon dioxide \( \mathrm{CO}_2 \) and water \( \mathrm{H}_2\mathrm{O} \) when a hydrocarbon combusts. For example, in our reaction:

• Acetylene \( \mathrm{C}_2\mathrm{H}_2 \) combines with oxygen \( \mathrm{O}_2 \).
• The products are \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \).

Think of a combustion reaction like a firework show, where oxygen fuels the reaction, producing energy and flames.
These reactions are exothermic, meaning they release energy. This is why they are useful in engines and heaters. The balanced \(2 \mathrm{C}_2\mathrm{H}_2 + 5 \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 2 \mathrm{H}_2\mathrm{O} \) shows the ratio needed for combustion.

Limiting Reactant

The limiting reactant in a chemical reaction is like the weakest link in a chain. It limits the amount of product formed.
In our example, the limiting reactant is the substance that is completely consumed first:

• Compare the mole ratio from the balanced equation to the moles available.
• The balanced equation tells us for every 2 moles of \( \mathrm{C}_2\mathrm{H}_2 \), 5 moles of \( \mathrm{O}_2 \) are needed.
• We calculated 0.313 moles of \( \mathrm{O}_2 \), but 0.963 moles are needed to combust all the \( \mathrm{C}_2\mathrm{H}_2 \).

Thus, \( \mathrm{O}_2 \) is the limiting reactant because it runs out first, stopping the reaction from continuing further.

Mole Calculations

Mole calculations help you understand the quantity relationships between reactants and products. Moles are to chemistry what dozen is to counting eggs.
To calculate moles, you divide the mass of a substance by its molar mass. For example:

• For acetylene \( \mathrm{C}_2\mathrm{H}_2 \): \( \frac{10.0 \mathrm{~g}}{26 \mathrm{~g/mol}} \approx 0.385 \mathrm{~mol} \)
• For oxygen \( \mathrm{O}_2 \): \( \frac{10.0 \mathrm{~g}}{32 \mathrm{~g/mol}} \approx 0.313 \mathrm{~mol} \)

Using these values, you can determine how reactants are used and how much product is formed. It's an essential step for stoichiometry calculations.

Stoichiometry

Stoichiometry is the mathematical heart of chemistry that describes how elements and compounds react
in specific ratios. It can tell us "how much" in a reaction. With stoichiometry, you translate moles from reactants to products:

• Using the balanced equation, relate moles of reactants to moles of products.
• In our problem, moles of \( \mathrm{O}_2 \) determine the moles of \( \mathrm{CO}_2 \) and \( \mathrm{H}_2\mathrm{O} \) produced.

The heart of stoichiometry lies in the balanced chemical equation. From here, you can calculate how much product you expect if, say, \(0.313 \mathrm{~mol} \mathrm{O}_2\) completely reacts:

• \(0.313 \mathrm{~mol} \mathrm{O}_2 \) uses \(0.125 \mathrm{~mol} \mathrm{C}_2\mathrm{H}_2 \), based on the ratio from the equation (2:5).
• produces \(0.25 \mathrm{~mol} \mathrm{CO}_2 \) and \(0.25 \mathrm{~mol} \mathrm{H}_2\mathrm{O} \).

Stoichiometry not only guides you from reactants to products, but helps you manage resources efficiently in labs and industry.