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?

Short answer

The balanced chemical equation for the combustion of acetylene and oxygen is $2{\mathrm{C}}_2{\mathrm{H}}_2+5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$. The limiting reactant is oxygen (${\mathrm{O}}_2$). After the reaction, the masses of ${\mathrm{C}}_2{\mathrm{H}}_2$, ${\mathrm{O}}_2$, ${\mathrm{CO}}_2$, and ${\mathrm{H}}_2\mathrm{O}$ remaining are: $0\mathrm{g}$, $0\mathrm{g}$, $55.01\mathrm{g}$, and $11.26\mathrm{g}$, respectively.

Step-by-step solution

Write the unbalanced chemical equation

The combustion of acetylene (${\mathrm{C}}_2{\mathrm{H}}_2$) in the presence of oxygen (${\mathrm{O}}_2$) produces carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$). The unbalanced chemical equation can be written as: $${\mathrm{C}}_2{\mathrm{H}}_2+{\mathrm{O}}_2\to {\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$$

Balance the chemical equation

To balance the chemical equation, we adjust the coefficients of the reactants and products such that the number of atoms of each element is equal on both sides of the equation. The balanced chemical equation is: $$2{\mathrm{C}}_2{\mathrm{H}}_2+5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$$

Calculate the moles of reactants

Given the mass of each reactant (10.0 g), we can calculate the moles of acetylene and oxygen. The molar mass of ${\mathrm{C}}_2{\mathrm{H}}_2$ is: $2\times 12.01\mathrm{g}/\mathrm{mol}+2\times 1.01\mathrm{g}/\mathrm{mol}=26.04\mathrm{g}/\mathrm{mol}$ The molar mass of ${\mathrm{O}}_2$ is: $2\times 16.00\mathrm{g}/\mathrm{mol}=32.00\mathrm{g}/\mathrm{mol}$ The moles of acetylene are: $\frac{10.0\mathrm{g}}{26.04\mathrm{g}/\mathrm{mol}}=0.384\mathrm{mol}$ The moles of oxygen are: $\frac{10.0\mathrm{g}}{32.00\mathrm{g}/\mathrm{mol}}=0.3125\mathrm{mol}$

Determine the limiting reactant

From the balanced equation, we see that 2 moles of ${\mathrm{C}}_2{\mathrm{H}}_2$ react with 5 moles of ${\mathrm{O}}_2$. To compare the moles of each reactant, we must calculate the mole ratios. The mole ratio of acetylene is: $\frac{0.384\mathrm{mol}}{2}=0.192$ The mole ratio of oxygen is: $\frac{0.3125\mathrm{mol}}{5}=0.0625$ Since the mole ratio of oxygen ($0.0625$) is less than that of acetylene ($0.192$), oxygen (${\mathrm{O}}_2$) is the limiting reactant.

Calculate the mass of unreacted and formed substances

First, we will calculate the moles of consumed acetylene: $${\mathrm{consumed}\mathrm{C}}_2{\mathrm{H}}_2=2\times {\text{(moles of consumed O}}_2\text{)}=2\times 0.3125\mathrm{mol}=0.625\mathrm{mol}$$ Moles of unreacted acetylene: ${\mathrm{C}}_2{\mathrm{H}}_2=0.384-0.625=-0.241\mathrm{mol}$, the negative sign indicates that all the acetylene was consumed. All acetylene is consumed, which means there are 0 grams remaining after the reaction. Moles of unreacted oxygen: ${\mathrm{O}}_2=0-0.3125=-0.3125\mathrm{mol}$, the negative sign indicates that all the oxygen was consumed. All oxygen is consumed, which means there are 0 grams remaining after the reaction. Moles of carbon dioxide formed: $4\times {\text{(moles of consumed O}}_2\text{)}=4\times 0.3125\mathrm{mol}=1.25\mathrm{mol}$ Mass of carbon dioxide formed: $1.25\mathrm{mol}\times 44.01\mathrm{g}/\mathrm{mol}=55.01\mathrm{g}$ Moles of water formed: $2\times {\text{(moles of consumed O}}_2\text{)}=2\times 0.3125\mathrm{mol}=0.625\mathrm{mol}$ Mass of water formed: $0.625\mathrm{mol}\times 18.02\mathrm{g}/\mathrm{mol}=11.26\mathrm{g}$ So, the masses of ${\mathrm{C}}_2{\mathrm{H}}_2$, ${\mathrm{O}}_2$, ${\mathrm{CO}}_2$, and ${\mathrm{H}}_2\mathrm{O}$ remaining after the reaction are: $${\mathrm{C}}_2{\mathrm{H}}_2=0\mathrm{g}$$ $${\mathrm{O}}_2=0\mathrm{g}$$ $${\mathrm{CO}}_2=55.01\mathrm{g}$$ $${\mathrm{H}}_2\mathrm{O}=11.26\mathrm{g}$$

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant determines how much product is formed. It's the substance that gets used up first, stopping the reaction because there’s nothing left of it to react. In our example, we compare the moles of each reactant. Using the balanced equation, we find that 2 moles of acetylene (${\mathrm{C}}_2{\mathrm{H}}_2$) react with 5 moles of oxygen (${\mathrm{O}}_2$).

• Calculate the moles of each reactant from their given masses.
• Determine the ratio of moles required by the balanced equation.
• Compare actual moles available to find the limiting reactant.

In this case, oxygen is the limiting reactant because it has the lower mole ratio, meaning it runs out first. Understanding limiting reactants is crucial in stoichiometry as it affects the amount of products formed.

Balanced Chemical Equation

A balanced chemical equation ensures that the same number of each type of atom appears on both sides of the equation. This is necessary to satisfy the law of conservation of mass, which states that matter can neither be created nor destroyed. The initial step involves writing an unbalanced equation for the reaction.

• Identify the reactants and products from the chemical reaction description.
• Adjust coefficients to balance the number of atoms on both sides.

For the combustion of acetylene, the balanced equation is:$$2{\mathrm{C}}_2{\mathrm{H}}_2+5{\mathrm{O}}_2\to 4{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$$Balancing allows us to correctly calculate the reactants and products in any stoichiometric calculations.

Mole Calculations

Mole calculations are essential steps in stoichiometry, allowing us to convert masses into moles and vice versa. This conversion uses the substance's molar mass, which is the mass of one mole of that substance. Here's how we perform mole calculations in this context:

• Calculate the molar mass of each reactant and product.
• Convert given masses into moles using the formula: $\text{moles}=\frac{\text{mass}}{\text{molar mass}}$.

For acetylene (${\mathrm{C}}_2{\mathrm{H}}_2$) and oxygen (${\mathrm{O}}_2$), these calculations convert 10.0 g of each into moles, revealing how much of each substance participates in the reaction. Mastering these conversions helps solve complex stoichiometry problems and predict product amounts formed in a reaction.

Combustion Reaction

Combustion reactions involve a substance reacting with oxygen to produce heat and light, often resulting in the formation of oxides like carbon dioxide (${\mathrm{CO}}_2$) and water (${\mathrm{H}}_2\mathrm{O}$). These reactions are a common example of oxidation-reduction processes. In our example, acetylene combusts in oxygen. Here's what happens:

• The reactants are acetylene and oxygen.
• The products are carbon dioxide and water.
• The reaction releases energy in the form of heat and light.

Combining these reactions with stoichiometric concepts helps us understand real-world applications, such as how fuel burns in engines, powering daily activities.