Question

Cyanogen gas, \(\mathrm{C}_{2} \mathrm{~N}_{2}\), has been found in the gases of outer space. It can react with fluorine to form carbon tetrafluoride and nitrogen trifluoride. $$ \mathrm{C}_{2} \mathrm{~N}_{2}(g)+7 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+2 \mathrm{NF}_{3}(g) $$ (a) How many moles of fluorine react with \(1.37 \mathrm{~mol}\) of cyanogen? (b) How many moles of \(\mathrm{CF}_{4}\) are obtained from \(13.75 \mathrm{~mol}\) of fluorine? (c) How many moles of cyanogen are required to produce \(0.8974 \mathrm{~mol}\) of \(\mathrm{NF}_{3}\) ? (d) How many moles of fluorine will yield \(4.981 \mathrm{~mol}\) of nitrogen trifluoride?

Short answer

b) How many moles of CF4 are obtained from 13.75 moles of fluorine (F2)? c) How many moles of cyanogen are required to produce 0.8974 moles of nitrogen trifluoride (NF3)? d) How many moles of fluorine will yield 4.981 moles of nitrogen trifluoride (NF3)? Answer: a) 9.59 moles of fluorine b) 3.93 moles of CF4 c) 0.4487 moles of cyanogen d) 17.43 moles of fluorine

Step-by-step solution

Determine moles of cyanogen

Given, 1.37 moles of cyanogen.

Find the mole ratio of cyanogen to fluorine from the balanced equation

From the balanced equation, 1 mole of C2N2 reacts with 7 moles of F2.

Calculate moles of fluorine required

Using the mole ratio, determine the moles of fluorine required for the reaction with 1.37 moles of cyanogen: (1.37 moles C2N2) x (7 moles F2 / 1 mole C2N2) = 9.59 moles F2. Answer (a): 9.59 moles of fluorine react with 1.37 moles of cyanogen. #PART B#

Determine moles of fluorine

Given, 13.75 moles of fluorine.

Find the mole ratio of fluorine to CF4 from the balanced equation

From the balanced equation, 7 moles of F2 produce 2 moles of CF4.

Calculate moles of CF4 obtained

Using the mole ratio, determine the moles of CF4 obtained from 13.75 moles of fluorine: (13.75 moles F2) x (2 moles CF4 / 7 moles F2) = 3.93 moles CF4. Answer (b): 3.93 moles of CF4 are obtained from 13.75 moles of fluorine. #PART C#

Determine moles of NF3

Given, 0.8974 moles of NF3.

Find the mole ratio of cyanogen to NF3 from the balanced equation

From the balanced equation, 1 mole of C2N2 produces 2 moles of NF3.

Calculate moles of cyanogen required

Using the mole ratio, determine the moles of cyanogen required to produce 0.8974 moles of NF3: (0.8974 moles NF3) x (1 mole C2N2 / 2 moles NF3) = 0.4487 moles C2N2. Answer (c): 0.4487 moles of cyanogen are required to produce 0.8974 moles of NF3. #PART D#

Determine moles of NF3

Given, 4.981 moles of NF3.

Find the mole ratio of fluorine to NF3 from the balanced equation

From the balanced equation, 7 moles of F2 produce 2 moles of NF3.

Calculate moles of fluorine required

Using the mole ratio, determine the moles of fluorine required to yield 4.981 moles of NF3: (4.981 moles NF3) x (7 moles F2 / 2 moles NF3) = 17.43 moles F2. Answer (d): 17.43 moles of fluorine will yield 4.981 moles of nitrogen trifluoride.

Key concepts

Mole Ratio

Understanding the mole ratio in chemical reactions is a cornerstone of stoichiometry, the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. The mole ratio, derived from the coefficients of a balanced chemical equation, showcases the proportional relationship between the quantities of all substances involved.

For instance, in the reaction between cyanogen and fluorine to produce carbon tetrafluoride and nitrogen trifluoride, the balanced equation tells us that for every one mole of cyanogen, seven moles of fluorine are required. This 1:7 mole ratio is crucial for calculations, as it allows us to predict how much of one substance will react or be produced from a given amount of another substance.

When solving stoichiometry problems, it's essential to interpret the balanced equation correctly to ascertain the mole ratios. These ratios serve as conversion factors between moles of reactants and products, thereby enabling us to solve problems such as determining the amount of reactants required or products formed. By utilizing these mole ratios accurately, one can make precise predictions about the outcomes of chemical reactions.

Chemical Reaction Calculations

Performing chemical reaction calculations is an integral part of stoichiometry that allows us to determine the amounts of substances consumed or produced in a chemical reaction. This typically involves three key steps: identifying the mole ratios, using them as conversion factors, and applying dimensional analysis to compute the quantities.

The process begins with a balanced chemical equation, which establishes the framework for the calculations. Then, using the mole ratios determined from the equation, you can calculate the moles of reactants needed or the moles of products that will be formed. In practice, this often involves cross-multiplication and division to convert moles of one substance to moles of another based on the ratios provided by the balanced equation. For improved clarity: if we know the amount of one substance in a reaction, we can find the amount of any other substance in the reaction using the mole ratio.

During instruction, it's beneficial to emphasize the importance of unit consistency and conversions, ensuring students can successfully navigate between grams, moles, and particles (such as atoms or molecules), which is often required in stoichiometric calculations. Tools such as the periodic table and Avogadro's number play pivotal roles in such conversions. Simplifying these complex concepts through methodical step-by-step solutions and emphasizing their real-world applications can greatly enhance students’ understanding.

Balanced Chemical Equations

A balanced chemical equation is essential for accurately predicting the amounts of reactants and products involved in a chemical reaction. It respects the law of conservation of mass and ensures that the number of atoms of each element is the same on both sides of the reaction.

To balance a chemical equation, one must adjust the coefficients (the numbers in front of the chemical formulas) to achieve the same number of atoms for each element in both reactants and products. For example, the equation for the reaction of cyanogen with fluorine to form carbon tetrafluoride and nitrogen trifluoride is carefully balanced so that the atoms of carbon, nitrogen, and fluorine tally on both sides of the reaction.

Students should learn the significance of a balanced equation as it impacts all quantitative aspects of the reaction, such as mole ratios and ultimately, stoichiometric calculations. By solidly grounding students in the method of balancing equations, they will be better equipped for subsequent tasks in stoichiometry, including real-world applications such as determining the quantity of a reactant required for a reaction or calculating product yields, which are vital skills in both academic studies and various scientific industries.