Problem 99 - Chemistry Matter and Change

Question

Fertilizer The compound calcium cyanamide \((\) CaNCN) is used as a nitrogen source for crops. To obtain this compound, calcium carbide is reacted with nitrogen at high temperatures. \begin{equation} \mathrm{CaC}_{2}(\mathrm{s})+\mathrm{N}_{2}(\mathrm{g}) \rightarrow \mathrm{CaNCN}(\mathrm{s})+\mathrm{C}(\mathrm{s}) \end{equation} What mass of CaNCN can be produced if 7.50 mol of CaC \(_{2}\) reacts with 5.00 mol of \(N_{2} ?\)

Step-by-Step Solution

Answer

400.50 grams of CaNCN can be produced.

Step 1: Identify the Limiting Reactant

To find the limiting reactant, compare the stoichiometric ratio from the balanced equation with the given moles of reactants. The reaction is: \( \text{CaC}_2 (s) + \text{N}_2 (g) \rightarrow \text{CaNCN} (s) + \text{C} (s) \). The stoichiometric ratio is 1:1 (1 mole of \(\text{CaC}_2\) reacts with 1 mole of \(\text{N}_2\)). Given 7.50 moles of \(\text{CaC}_2\) and 5.00 moles of \(\text{N}_2\), \(\text{N}_2\) is the limiting reactant since there is less than the required amount of \(\text{CaC}_2\).

Step 2: Calculate Moles of CaNCN Produced

With \(\text{N}_2\) as the limiting reactant, the reaction will completely consume 5.00 moles of \(\text{N}_2\). According to the balanced equation, 1 mole of \(\text{N}_2\) produces 1 mole of \(\text{CaNCN}\). Therefore, 5.00 moles of \(\text{N}_2\) will produce 5.00 moles of \(\text{CaNCN}\).

Step 3: Calculate Mass of CaNCN Produced

The molar mass of \( \text{CaNCN} \) needs to be calculated first. \( \text{Ca} \) has a molar mass of 40.08 g/mol, \( \text{N} \) has a molar mass of 14.01 g/mol, and \( \text{C} \) has a molar mass of 12.01 g/mol. Add them together: \( 40.08 + 14.01 + 2\times 12.01 = 80.10 \ g/mol \). Multiply the molar mass by the moles of \( \text{CaNCN} \) to find the mass: \( 5.00 \text{ moles} \times 80.10 \text{ g/mol} = 400.50 \text{ grams} \).

Key Concepts

Limiting Reactant

In stoichiometry, identifying the limiting reactant is a crucial step to determine how much product can be formed in a chemical reaction. The limiting reactant is the reactant that is completely consumed first, stopping the reaction from continuing. It effectively limits the amount of products that can be formed.
To find out which reactant is the limiting one, compare the amount of reactants used in the reaction with their stoichiometric ratios derived from the balanced chemical equation.
If the equation shows that 1 mole of Reactant A reacts with 1 mole of Reactant B, and you have 3 moles of A and 2 moles of B, then B is the limiting reactant because there is not enough of it to fully react with all of A according to the 1:1 ratio.
In our exercise, based on the balanced reaction formula, the stoichiometric ratio between CaC₂ and N₂ is 1:1. Given 7.50 moles of CaC₂ and 5.00 moles of N₂, nitrogen (N₂) becomes the limiting reactant due to its lower mole count compared to the 1:1 requirement.

Molar Mass Calculation

Calculating the molar mass is essential to converting moles of a compound to grams, which is often necessary in stoichiometry to find out how much product can be made or how much reactant will be used.
The molar mass is the sum of the mass of all atoms in a single molecule of a compound. You find it by adding the atomic masses of all the elements in the compound. Atomic masses are usually found on the periodic table.

• For calcium cyanamide (CaNCN), the molar mass calculation involves:
• Calcium (Ca): 40.08 g/mol
• Nitrogen (N): 14.01 g/mol
• Carbon (C): 12.01 g/mol

Two carbon atoms are present, so their masses are doubled. The computation is:
\[40.08 + 14.01 + (2 imes 12.01) = 80.10 \text{ g/mol} \]By knowing the moles involved and the molar mass, converting to grams is straightforward. Multiply the molar mass by the moles to calculate the mass of the compound.

Balanced Chemical Equation

A balanced chemical equation accurately represents all reactants and products in a reaction. This is essential for correctly applying stoichiometry, as it ensures the law of conservation of mass is obeyed.
Every chemical equation must be balanced, meaning the number and type of atoms are the same on both sides of the equation.

• This is achieved by adjusting the coefficients (numbers before molecules), not by changing the chemical formulas of the reactants or products.
• The coefficients provide the ratios of moles necessary for the reaction.

In our fertilizer example:\[\text{CaC}_2(s) + \text{N}_2(g) \rightarrow \text{CaNCN}(s) + \text{C}(s) \]This shows a 1:1:1:1 ratio, meaning one mole of calcium carbide reacts with one mole of nitrogen to produce one mole of calcium cyanamide and one mole of carbon.
Understanding how to balance chemical equations is fundamental to solving moles, masses, and yields of reactions correctly and ensures you're working with accurate data for the calculations involved.