Question

When a mixture of 10.0 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?

Short answer

The balanced chemical equation for the combustion of acetylene (C2H2) and oxygen (O2) is \( 2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} + 2H_{2}O \). The limiting reactant is O2. After the reaction is complete, there are 7.00g of C2H2, 0g of O2, 11.0g of CO2, and 2.25g of H2O present.

Step-by-step solution

Write the balanced chemical equation.

The combustion reaction between acetylene (C2H2) and oxygen (O2) produces CO2 and H2O. So, the unbalanced equation is: \( C_{2}H_{2} + O_{2} \rightarrow CO_{2} + H_{2}O \) Now, balance the equation: 2 C2H2 + 5 O2 → 4 CO2 + 2 H2O The balanced chemical equation is: \( 2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} + 2H_{2}O \)

Calculate the moles of each reactant.

Given masses: 10.0 g of C2H2 and 10.0 g of O2 Molar mass of C2H2 = (2 × 12.01) + (2 × 1.01) = 26.04 g/mol Molar mass of O2 = 2 × 16.00 = 32.00 g/mol Now, we can calculate the moles of each reactant: Moles of C2H2 = (10.0 g) / (26.04 g/mol) = 0.384 moles Moles of O2 = (10.0 g) / (32.00 g/mol) = 0.313 moles

Determine the limiting reactant.

Divide the moles of each reactant by their stoichiometric coefficients: C2H2: 0.384 moles / 2 = 0.192 O2: 0.313 moles / 5 = 0.063 The lower value is for O2, so oxygen is the limiting reactant.

Calculate the moles of each product formed and remaining moles of excess reactant.

Since oxygen is the limiting reactant, use its moles to find the moles of products formed: Moles of CO2 = (4/5) × 0.313 moles = 0.250 moles Moles of H2O = (2/5) × 0.313 moles = 0.125 moles And the moles of C2H2 remaining after the reaction: Moles of C2H2 remaining = 0.384 moles - (2 × 0.313 moles / 5) = 0.269 moles

Calculate the mass of each compound after the reaction is complete.

Convert the moles of each compound to mass: Mass of C2H2 remaining = 0.269 moles × 26.04 g/mol = 7.00 g Mass of O2 remaining = 0 g (all used up) Mass of CO2 formed = 0.250 moles × 44.01 g/mol = 11.0 g Mass of H2O formed = 0.125 moles × 18.02 g/mol = 2.25 g After the reaction is complete, there are 7.00g of C2H2, 0g of O2, 11.0g of CO2, and 2.25g of H2O present.

Key concepts

Balanced Chemical Equation

Understanding the balanced chemical equation is a crucial aspect of chemistry, especially when studying combustion reactions like that of acetylene. A balanced equation ensures that the law of conservation of mass is adhered to, meaning that the number of atoms of each element is the same on both sides of the equation. For the acetylene combustion, we balance the initial unbalanced equation, getting \[\begin{equation}\2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} + 2H_{2}O\end{equation}\]This process requires knowing the proper stoichiometric coefficients, which are the numbers placed before each compound (2 for C2H2, 5 for O2, etc.). These coefficients indicate the precise ratios in which reactants combine and products form, ensuring the balance of atoms.

Limiting Reactant

The concept of the limiting reactant is akin to the bottleneck in a production process—it determines the amount of product that can be formed in a chemical reaction. It is the reactant that will be consumed first, thus limiting the extent of the reaction. Identifying it requires comparing the number of moles of the reactants with the stoichiometric ratios from the balanced equation. In our acetylene combustion exercise, we determined that oxygen (O2) is the limiting reactant because it has the smallest ratio when its moles are divided by the stoichiometric coefficient. This determination is fundamental because it helps predict the quantity of each product formed.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It involves calculations based on the balanced equation, and it allows chemists to predict the amounts of substances consumed and produced. When we know the moles of the limiting reactant, we can use stoichiometric factors from the balanced equation to calculate the amount of products that will form, as demonstrated in the provided solution. This can also be reversed to find required amounts of reactants to produce a desired amount of product.

Mole Concept

The mole concept is indispensable when dealing with chemical reactions. A mole represents Avogadro's number (\[\begin{equation}6.022 \times 10^{23}\end{equation}\]) of particles (atoms, molecules, ions, etc.), which is akin to a dozen representing 12 items. By converting mass to moles using the molar mass, we can easily utilize the balanced equation, which is expressed in moles rather than grams. In our exercise, we used molar mass to convert 10.0 g of acetylene and oxygen into moles before proceeding with the reaction stoichiometry.

Molar Mass Calculation

The molar mass calculation is a method to convert between the mass of a substance and the amount in moles. It is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). For example, the molar mass of acetylene (C2H2) is calculated by summing the molar masses of carbon and hydrogen in the molecule. This value is critical for converting the given mass of a substance to moles, as shown in the exercise. Knowing the molar mass of the reactants allowed us to find their initial moles, leading us to identify the limiting reactant and the amount of products formed.