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 \(1.37\) 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 \(\mathrm{NP}_{3} ?\) (d) How many moles of fluorine will yield \(4.981 \mathrm{~mol}\) of nitrogen trifluoride?

Short answer

Additionally, how many moles of cyanogen are required to produce 0.8974 moles of NF3, and how many moles of fluorine will yield 4.981 moles of nitrogen trifluoride? Answer: 9.59 moles of fluorine are required to react with 1.37 moles of cyanogen, and 3.9286 moles of CF4 are obtained from 13.75 moles of fluorine. Furthermore, 0.8974 moles of cyanogen are required to produce 0.8974 moles of NF3, and 17.4335 moles of fluorine will yield 4.981 moles of nitrogen trifluoride.

Step-by-step solution

Identify the mole ratio

Using the balanced chemical equation, we can see that the mole ratio of \(\mathrm{C}_{2}\mathrm{N}_{2}\) to \(\mathrm{F}_{2}\) is 1:7.

Calculate moles of fluorine

Using the mole ratio, we can determine the moles of fluorine needed to react with 1.37 mol of cyanogen: Moles of fluorine = 1.37 mol \(\mathrm{C}_{2}\mathrm{N}_{2} \times \frac{7 \mathrm{~mol} \mathrm{F}_{2}}{1 \mathrm{~mol} \mathrm{C}_{2}\mathrm{N}_{2}} = 9.59 \mathrm{~mol} \mathrm{F}_{2}\) So, 9.59 moles of fluorine react with 1.37 moles of cyanogen. (b) How many moles of \(\mathrm{CF}_{4}\) are obtained from 13.75 mol of fluorine?

Identify the mole ratio

Using the balanced chemical equation, we can see that the mole ratio of \(\mathrm{F}_{2}\) to \(\mathrm{CF}_{4}\) is 7:2.

Calculate moles of \(\mathrm{CF}_{4}\)

Using the mole ratio, we can determine the moles of \(\mathrm{CF}_{4}\) obtained from 13.75 mol of fluorine: Moles of \(\mathrm{CF}_{4}\) = 13.75 mol \(\mathrm{F}_{2} \times \frac{2 \mathrm{~mol} \mathrm{CF}_{4}}{7 \mathrm{~mol} \mathrm{F}_{2}} = 3.9286 \mathrm{~mol} \mathrm{CF}_{4}\) So, 3.9286 moles of \(\mathrm{CF}_{4}\) are obtained from 13.75 moles of fluorine. (c) How many moles of cyanogen are required to produce 0.8974 mol of \(\mathrm{NF}_{3}\)?

Identify the mole ratio

Using the balanced chemical equation, we can see that the mole ratio of \(\mathrm{C}_{2}\mathrm{N}_{2}\) to \(\mathrm{NF}_{3}\) is 1:1.

Calculate moles of cyanogen

Using the mole ratio, we can determine the moles of cyanogen required to produce 0.8974 mol of \(\mathrm{NF}_{3}\): Moles of cyanogen = 0.8974 mol \(\mathrm{NF}_{3} \times \frac{1 \mathrm{~mol} \mathrm{C}_{2}\mathrm{N}_{2}}{1 \mathrm{~mol} \mathrm{NF}_{3}} = 0.8974 \mathrm{~mol} \mathrm{C}_{2}\mathrm{N}_{2}\) So, 0.8974 moles of cyanogen are required to produce 0.8974 moles of \(\mathrm{NF}_{3}\). (d) How many moles of fluorine will yield 4.981 mol of nitrogen trifluoride?

Identify the mole ratio

Using the balanced chemical equation, we can see that the mole ratio of \(\mathrm{F}_{2}\) to \(\mathrm{NF}_{3}\) is 7:2.

Calculate moles of fluorine

Using the mole ratio, we can determine the moles of fluorine needed to yield 4.981 mol of \(\mathrm{NF}_{3}\): Moles of fluorine = 4.981 mol \(\mathrm{NF}_{3} \times \frac{7 \mathrm{~mol} \mathrm{F}_{2}}{2 \mathrm{~mol} \mathrm{NF}_{3}} = 17.4335 \mathrm{~mol} \mathrm{F}_{2}\) So, 17.4335 moles of fluorine will yield 4.981 moles of nitrogen trifluoride.

Key concepts

Chemical Reactions

A chemical reaction is a process where substances, known as reactants, undergo a transformation to form different substances, called products. In a balanced chemical equation, the number of atoms for each element is the same on the reactant side as on the product side. This reflects the conservation of mass.

For example, in the given exercise, cyanogen gas reacts with fluorine to produce carbon tetrafluoride and nitrogen trifluoride. This is represented by the chemical equation \(\mathrm{C}_{2}\mathrm{N}_{2}(g) + 7\mathrm{F}_{2}(g) \longrightarrow 2\mathrm{CF}_{4}(g) + 2\mathrm{NF}_{3}(g)\). Here, the coefficients indicate the relative number of moles of each reactant and product needed to balance the reaction.

Understanding chemical reactions is fundamental in stoichiometric calculations, as it allows us to predict the amounts of products formed from given reactants.

Mole Ratio

The mole ratio is a critical concept in stoichiometry used to calculate the amounts of reactants and products in a chemical reaction. It is derived from the balanced chemical equation and represents the proportion of moles of one substance to the moles of another substance.

In our exercise example, the balanced equation yields mole ratios such as \(1:7\) for \(\mathrm{C}_{2}\mathrm{N}_{2}\) to \(\mathrm{F}_{2}\), and \(7:2\) for \(\mathrm{F}_{2}\) to \(\mathrm{CF}_{4}\) and \(\mathrm{NF}_{3}\). These ratios are pivotal to solving stoichiometric problems. Knowing the mole ratio, one can easily calculate how many moles of one chemical are needed to react with a given amount of another chemical, which was shown in the step by step solution where the mole ratios were applied to find the corresponding moles of fluorine and carbon tetrafluoride for given moles of cyanogen and nitrogen trifluoride, respectively.

Gas Reactions

Gas reactions are chemical reactions involving at least one gaseous reactant or product. They follow the same stoichiometric principles as reactions in other states, but with additional considerations for gas laws when calculating volumes, pressures, and temperatures.

For instance, in the exercise's reaction involving cyanogen and fluorine, both are gases, and the products formed are also gases. It's essential to account for the fact that gases can expand and contract under different conditions, which can affect stoichiometry under non-standard conditions. However, in a theoretical setup or under standard conditions (usually defined as 0°C and 1 atm pressure), we typically treat gases ideally and apply stoichiometry as we do for solids or liquids.

Standard Conditions

In many stoichiometric calculations for gases, we assume that the reaction occurs under standard temperature and pressure (STP). At STP, one mole of an ideal gas occupies 22.4 L, allowing us to convert between moles and volume easily if needed.