Question

The following demonstration takes place in a two-step process: First, solid calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with liquid water to produce acetylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and aqueous calcium hydroxide. Second, the acetylene gas produced is then ignited with a match, causing the combustion reaction of acetylene with oxygen gas to produce gaseous carbon dioxide and gaseous water. a. Write the balanced equations for each reaction that is occurring, including all phases. b. If a 100.0 -g sample of calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) is initially reacted with \(50.0 \mathrm{~g}\) of water, which reactant is limiting? c. Now imagine that the final gases produced are collected in a large balloon and allowed to cool to room temperature. Using the information from part b ( \(100.0 \mathrm{~g}\) of \(\mathrm{CaC}_{2}\) reacting with \(50.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) ), how many liters of carbon dioxide gas were produced in the balloon at a pressure of 1.00 atm and \(25^{\circ} \mathrm{C} ?\)

Short answer

a. The balanced equations for each reaction are: \( CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2 (g) + Ca(OH)_2(aq) \) and \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g) \) b. Water (H2O) is the limiting reactant. c. The volume of carbon dioxide gas (CO2) produced in the balloon is approximately 67.7 L.

Step-by-step solution

Write Balanced Equations

We have two reactions occurring. First, solid calcium carbide (CaC2) reacts with liquid water to produce acetylene gas (C2H2) and aqueous calcium hydroxide. The second reaction is the combustion reaction of the acetylene gas produced with oxygen gas to produce carbon dioxide gas and water gas. First Reaction: \( CaC_2(s) + 2H_2O(l) \rightarrow C_2H_2 (g) + Ca(OH)_2(aq) \) Second Reaction: \( 2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g) \)

Determine the Limiting Reactant

We have 100.0 g of CaC2 and 50.0 g of H2O. To find out which reactant is limiting, we should compare the moles of each reactant consumed in the reaction. First, calculate the molar mass of CaC2 and H2O: Molar mass of CaC2: \( 40.08 g/mol (Ca) + 2×12.01 g/mol (C) = 64.10 g/mol \) Molar mass of H2O: \( 2×1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol \) Now, calculate the moles of each reactant: Moles of CaC2 = \( \frac{100.0 g}{64.10 g/mol} = 1.56 mol \) Moles of H2O = \( \frac{50.0 g}{18.02 g/mol} = 2.77 mol \) Compare the mole ratio of reactants with the stoichiometry of the first balanced equation: Mole ratio: \( \frac{1.56 mol}{2.77mol} = 0.56 \) Stoichiometric ratio: \( \frac{1}{2} = 0.5 \) Since the mole ratio (0.56) is greater than the stoichiometric ratio (0.5), H2O is the limiting reactant.

Calculate the Volume of CO2 Produced

Now we can use the limiting reactant (H2O) to find out the amount of CO₂ produced. First, determine the moles of CO2 produced using the stoichiometry from the second balanced equation. Moles of CO2 = \( \frac{2}{2}\times moles~~of~~C_2H_2 \) But we know that moles of \(C_2H_2 = \frac{1}{2}\times moles~~of~~H_2O \) So, moles of CO2 = \( moles~~of~~H_2O = 2.77 mol \) Now that we have the moles of CO2, we can use the ideal gas law to find the volume of the gas at room temperature and pressure. \( PV = nRT \) \( V = \frac{nRT}{P} \) Where: \( V \) is volume, \( n \) is moles of CO2, \( R \) is the ideal gas constant \( \left( 0.0821 \frac{L \cdot atm}{mol \cdot K}\right) \), \( T \) is temperature in Kelvin, and \( P \) is pressure. To convert the temperature to Kelvin: Temperature in Kelvin: \( T = 25^\circ C + 273.15 = 298.15 K \) Now, calculate the volume of CO2 produced: \( V = \frac{2.77 mol × 0.0821 \frac{L \cdot atm}{mol \cdot K} × 298.15 K}{1.00 atm} \) \( V ≈ 67.7 L \) The volume of carbon dioxide gas produced in the balloon is approximately 67.7 L.

Key concepts

Limiting Reactant

In chemical reactions, the concept of a limiting reactant is crucial. It determines the maximum amount of product that can be formed. The limiting reactant is the substance that is completely consumed first, halting the reaction as there can be no further progress without it.

To find the limiting reactant, we first need to determine the number of moles of each reactant involved in the reaction. Then, compare the actual mole ratio of the reactants to the ratio required by the balanced chemical equation.

• If the actual mole ratio is greater than required, the reactant being consumed more slowly is in excess.
• If the actual mole ratio is less than required, that reactant runs out first; thus it is the limiting reactant.

In our example with calcium carbide ( CaC_2 ) and water ( H_2O ), after computing their respective moles, we find that water has a ratio above the stoichiometric ratio needed. Therefore, water is the limiting reactant in this scenario.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that connects the physical properties of gases. It combines several independent laws (Boyle's, Charles's, and Avogadro's laws) into a single equation:

\(PV = nRT\)

• \(P\) represents the pressure of the gas.
• \(V\) is the volume it occupies.
• \(n\) denotes the moles of gas.
• \(R\) is the ideal gas constant, which is \(0.0821 \frac{L \, atm}{mol \, K}\).
• \(T\) is the temperature in Kelvin.

To find the volume of a gas, rearrange the equation to \(V = \frac{nRT}{P}\).

This equation assumes that interactions between gas molecules are negligible and that the gas molecules occupy no volume, making it ideal for theoretical calculations. In practice, it provides a close approximation under normal conditions of temperature and pressure.

In this problem, after finding the moles of carbon dioxide produced, we used the Ideal Gas Law to calculate its volume at a specific temperature and pressure.

Stoichiometry

Stoichiometry is the area of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It is based on the balanced chemical equation and uses the concept of mole ratios derived from this equation.

A balanced chemical equation indicates the proportion of molecules that react and form products. In our scenario, the first reaction shows that one mole of calcium carbide reacts with two moles of water to produce acetylene and calcium hydroxide. The second reaction, combustion of acetylene, further illustrates how these reactants convert to carbon dioxide and water gas.

• This balanced approach allows us to determine the limiting reactant and the theoretical yield of the product.
• In essence, it provides the ratios needed to convert from moles of one substance to moles of another in a reaction.

For example, once we determined the limiting reactant, stoichiometry helped us predict the amount of \(CO_2\) produced, which was then used in conjunction with the Ideal Gas Law to find its volume. Stoichiometry is thus essential for planning reactions efficiently in both laboratory and industrial chemistry.