Problem 99 - Chemistry: The Central Science

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 resultant 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

The balanced chemical equation for the combustion reaction is \(\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\). The limiting reactant is oxygen. After the reaction is complete, there are 6.72 g of \(\mathrm{C}_{2} \mathrm{H}_{2}\), 0 g of \(\mathrm{O}_{2}\), 11.00 g of \(\mathrm{CO}_{2}\), and 2.27 g of \(\mathrm{H}_{2} \mathrm{O}\) present.

Step 1: Write the balanced chemical equation for the reaction between acetylene and oxygen

In this step, we will write the balanced chemical equation for the combustion reaction between acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)) and oxygen (\(\mathrm{O}_{2}\)) to produce carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)). The unbalanced equation is: \(\mathrm{C}_{2} \mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\) To balance the equation, start by balancing carbon and hydrogen atoms first, then adjust the oxygen atoms: \(\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\) The balanced equation is: \(\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\)

Step 2: Determine the limiting reactant

To determine the limiting reactant, we must first find the number of moles of acetylene and oxygen: Moles of acetylene: \(n = m / M = 10.0~\text{g} / 26.04~\text{g/mol} = 0.384~\text{mol}\) Moles of oxygen: \(n = m / M = 10.0~\text{g} / 32.00~\text{g/mol} = 0.313~\text{mol}\) Next, we will use the balanced equation to compare the mole ratios: Mole ratio of \(\ce{C2H2}\) to \(\ce{O2}\) from the balanced equation : \(2:5\) So, for every 2 moles of acetylene consumed, 5 moles of oxygen are needed. Now let's see if we have enough oxygen for the given moles of acetylene: Oxygen needed = \((0.384~\text{mol} \,\ce{C2H2}) \times (5~\text{mol} \,\ce{O2} / 2~\text{mol} \,\ce{C2H2}) = 0.960~\text{mol} \,\ce{O2}\) Since we only have 0.313 mol of oxygen, oxygen is the limiting reactant.

Step 3: Calculate the amounts of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) present after the reaction

Now that we've identified oxygen as the limiting reactant, we will calculate the moles of each species after the reaction, using the stoichiometry from the balanced equation: Moles of \(\ce{C2H2}\) consumed: \((0.313~\text{mol} \,\ce{O2}) \times (2~\text{mol} \,\ce{C2H2} / 5~\text{mol} \,\ce{O2}) = 0.126~\text{mol} \,\ce{C2H2}\) Moles of \(\ce{CO2}\) produced: \((0.313~\text{mol} \,\ce{O2}) \times (4~\text{mol} \,\ce{CO2} / 5~\text{mol} \,\ce{O2}) = 0.250~\text{mol} \,\ce{CO2}\) Moles of \(\ce{H2O}\) produced: \((0.313~\text{mol} \,\ce{O2}) \times (2~\text{mol} \,\ce{H2O} / 5~\text{mol} \,\ce{O2}) = 0.126~\text{mol} \,\ce{H2O}\) Now we will calculate the masses of each species present after the reaction: Mass of \(\ce{C2H2}\) remaining: \((0.384 - 0.126)~\text{mol} \times 26.04~\text{g/mol} = 6.72~\text{g}\) Mass of \(\ce{O2}\) remaining: \(0~\text{g}\) (All oxygen was consumed in the reaction) Mass of \(\ce{CO2}\) produced: \(0.250~\text{mol} \times 44.01~\text{g/mol} = 11.00~\text{g}\) Mass of \(\ce{H2O}\) produced: \(0.126~\text{mol} \times 18.02~\text{g/mol} = 2.27~\text{g}\) So after the reaction is complete, there are 6.72 g of \(\mathrm{C}_{2} \mathrm{H}_{2}\), 0 g of \(\mathrm{O}_{2}\), 11.00 g of \(\mathrm{CO}_{2}\), and 2.27 g of \(\mathrm{H}_{2} \mathrm{O}\) present.

Key Concepts

Stoichiometry

Stoichiometry is a central concept in chemistry that refers to the calculation of reactants and products in chemical reactions. It is based on the principle that in chemical reactions, matter is neither created nor destroyed. To understand stoichiometry, it is essential to know how to convert between mass and moles – a measure of the amount of substance. This is crucial because chemical equations use moles to showcase the proportions of reactants and products.

In the example of acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)) reacting with oxygen (\(\mathrm{O}_{2}\)), we first converted grams to moles using the molar mass of each substance. For acetylene and oxygen, the conversions were:

• Acetylene: Mass = 10.0 g, Molar Mass ≈ 26.04 g/mol, Moles ≈ 0.384 mol
• Oxygen: Mass = 10.0 g, Molar Mass = 32.00 g/mol, Moles ≈ 0.313 mol

These calculations set the stage for determining the limiting reactant – the substance that runs out first, limiting the amount of product formed. Understanding stoichiometry allows us to predict the quantities of products and leftover reactants after the reaction.

Chemical Equation Balancing

Balancing chemical equations ensures that the law of conservation of mass is upheld, meaning that the number of atoms for each element in the reactants equals the number in the products. This step is vital for accurately performing stoichiometric calculations and determining the reaction's limiting reactant.

For the combustion of acetylene and oxygen, the unbalanced equation starts as:

• \(\mathrm{C}_{2} \mathrm{H}_{2} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2} \mathrm{O}\)

To balance it:

• First, balance carbon atoms, then hydrogen, and finally oxygen. This sequence simplifies the process.
• Balanced Equation: \(\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\)

Here you see 2 moles of acetylene react with 5 moles of oxygen to produce 4 moles of carbon dioxide and 2 moles of water. Ensuring the equation is balanced is crucial because stoichiometric calculations depend on the mole ratio derived from it.

Combustion Reaction

A combustion reaction typically involves a substance reacting with oxygen to produce oxide products and release heat and light. These reactions are common in everyday life, like burning fuels in engines or fireplaces.

In the case of acetylene (\(\mathrm{C}_{2} \mathrm{H}_{2}\)), the combustion reaction produces carbon dioxide (\(\mathrm{CO}_{2}\)) and water (\(\mathrm{H}_{2} \mathrm{O}\)) as products according to the balanced chemical equation:

• \(\ce{2C2H2 + 5O2 -> 4CO2 + 2H2O}\)

During combustion, the hydrocarbons in acetylene oxidize, leading to the emission of energy.

This reaction specifically demonstrates complete combustion where all carbon transforms into carbon dioxide, and all hydrogen into water, assuming ample oxygen supply. When oxygen is limited, incomplete combustion may occur, producing carbon monoxide (\(\mathrm{CO}\)) or other by-products. Understanding combustion reactions is essential for applications in energy production, engineering, and environmental science.