Question

Calcium carbide, \(\mathrm{CaC}_{2},\) can be produced in an electric furnace by strongly heating calcium oxide (lime) with carbon. The unbalanced cquation is $$ \mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) $$ Calcium carbide is useful because it reacts readily with water to form the flammable gas acctylene, \(\mathrm{C}_{2} \mathrm{H}_{2},\) which is used extensively in the welding industry. The unbalanced equation is $$ \mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) $$ What mass of acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2},\) would be produced by complete reaction of \(3.75 \mathrm{~g}\) of calcium carbide?

Short answer

The mass of acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2},\) produced by the complete reaction of \(3.75\, \text{g}\) of calcium carbide is approximately \(1.52\,\text{g}\).

Step-by-step solution

1. Balance the first equation

To balance the given equation \(\mathrm{CaO}(s)+\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g)\), we can determine that the balanced equation is: $$ \mathrm{CaO}(s)+2\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) $$

2. Balance the second equation

To balance the given equation \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s)\), we can determine that the balanced equation is: $$ \mathrm{CaC}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) $$

3. Calculate the moles of calcium carbide

Next, we need to find the moles of calcium carbide from its given mass (\(3.75\,\text{g}\)). We will use the molar mass of calcium carbide, which can be found from the periodic table as: $$ \text{Molar mass of }\mathrm{CaC}_2 = (40.08 + 2 \times 12.01) \,\text{g/mol} = 64.10\,\text{g/mol} $$ Now, using the formula \(n = m / M\), where \(n\) is the moles, \(m\) is the mass, and \(M\) is the molar mass, we can find the moles of calcium carbide: $$ n_\mathrm{CaC_2} = \frac{3.75}{64.10} = 0.0585\,\text{moles} $$

4. Calculate the moles of acetylene produced

Now that we have the moles of calcium carbide, we will use the stoichiometry of the balanced equation to find the moles of acetylene produced: $$ \mathrm{CaC}_{2}(s)+2\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{2} \mathrm{H}_{2}(g)+\mathrm{Ca}(\mathrm{OH})_{2}(s) $$ Since there is a 1:1 stoichiometric ratio between calcium carbide and acetylene, the moles of acetylene produced will be the same as the moles of calcium carbide reacted: $$ n_\mathrm{C_{2}H_{2}} = n_\mathrm{CaC_2} = 0.0585\,\text{moles} $$

5. Calculate the mass of acetylene produced

Finally, we will use the moles of acetylene produced and its molar mass to find the mass of acetylene produced. The molar mass of acetylene is: $$ \text{Molar mass of }\mathrm{C}_{2} \mathrm{H}_{2} = (2 \times 12.01 + 2 \times 1.01) \,\text{g/mol} = 26.04\,\text{g/mol} $$ Using the formula \(m = n \times M\), where \(m\) is the mass, \(n\) is the moles, and \(M\) is the molar mass, we can find the mass of acetylene produced: $$ m_\mathrm{C_{2}H_{2}} = 0.0585 \times 26.04 = 1.52\,\text{g} $$ So, the mass of acetylene gas produced by the complete reaction of \(3.75\, \text{g}\) of calcium carbide is approximately \(1.52\,\text{g}\).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, transform into different substances, called products. These reactions are governed by the rearrangement of atoms. The initial and new substances have distinct properties. In our example, calcium oxide and carbon react to form calcium carbide and carbon monoxide. During the second reaction, calcium carbide reacts with water to produce acetylene gas and calcium hydroxide. In a chemical reaction, it's crucial to understand the nature of the reactants and products.

• Reactants: Starting materials in a reaction. Here, calcium oxide and carbon are reactants.
• Products: Substances formed as a result of the reaction. Calcium carbide and acetylene are products in the given equations.

Understanding chemical reactions allows for predicting the outcomes and implications in real-world applications like manufacturing and welding.

Balancing Equations

Balancing equations ensures the law of conservation of mass, stating that mass cannot be created or destroyed in a chemical reaction. Each side of a balanced equation must reflect the same number of atoms of each element.

Steps to Balance Equations

First, write the reactants and products with their chemical formulas. Next, adjust coefficients to have an equal number of each type of atom on both sides. For example, the initial unbalanced equation for the first reaction is \(\mathrm{CaO}(s) + \mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) \)We balanced it by adjusting the coefficient for carbon to "2":\(\mathrm{CaO}(s) + 2\mathrm{C}(s) \rightarrow \mathrm{CaC}_{2}(s) + \mathrm{CO}(g)\).Balancing involves trial and error until all elements are equal on both sides. This is repeated for the second equation in the original solution to reflect stoichiometry correctly.

Molar Mass

Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It bridges the concept of bulk material to a single molecule level, essential for stoichiometric calculations.

Calculating Molar Mass

To find the molar mass of a compound, sum the atomic masses of all atoms in the formula. For calcium carbide (\(\mathrm{CaC}_2\)), calculate as follows:- Calcium (Ca): atomic mass = 40.08 g/mol- Carbon (C): atomic mass = 12.01 g/molThus, the molar mass of \(\mathrm{CaC}_2\) is calculated as: \((40.08 + 2 \times 12.01)\,\text{g/mol} = 64.10\,\text{g/mol}\).Molar mass enables the conversion of mass to moles, crucial for determining the quantity of a substance and predicting product formation.

Gas Production

Gas production in chemical reactions can often be directly observed through bubbling or fizzing in solutions. In stoichiometric calculations, understanding gas production is key for determining yield and efficiency. Here, acetylene gas (\(\mathrm{C}_{2}\mathrm{H}_{2}\)) is produced from calcium carbide.

Stoichiometry and Gas Production

Gas-producing reactions are important in not only measuring reaction progress but also industrially beneficial processes, like welding. Stoichiometry allows for predicting how much gas will form from known reactant quantities, using the balanced equation to establish ratios.For example, in the equation\(\mathrm{CaC}_{2}(s) + 2\mathrm{H}_{2}\mathrm{O}(l) \rightarrow \mathrm{C}_{2}\mathrm{H}_{2}(g) + \mathrm{Ca(OH)}_{2}(s)\)We see a 1:1 stoichiometric relationship between calcium carbide and acetylene, meaning for every mole of calcium carbide, an equal mole of acetylene is produced. This allows prediction of exact amounts of gas obtained when reacting a known substance quantity.