Question

Boron nitride reacts with iodine monofluoride in trichlorofluoromethane at -30 ' \(\mathrm{C}\) to produce pure nitrogen triiodide and by-product (BF_). $$ \mathrm{BN}+3 \mathrm{IF} \rightarrow \mathrm{NI}_{3}+\mathrm{BF}_{3} $$ What mass of iodine monofluoride must be used to produce \(30.0 \mathrm{~g}\) of nitrogen triiodide? b. When \(30.0 \mathrm{~g}\) of nitrogen triiodide is produced, what is the maximum mass of by-product \(\left(\mathrm{BF}_{3}\right)\) created?

Short answer

To produce 30.0 g of nitrogen triiodide, approximately 33.23 g of iodine monofluoride must be used. When 30.0 g of nitrogen triiodide is produced, the maximum mass of by-product (BF_3) created is approximately 5.09 g.

Step-by-step solution

Calculate the molar masses of involved substances

First, we need to find the molar masses of BN, IF, NI_3 and BF_3. This can be determined using the periodic table and adding up the atomic weights of each element in the compound: Molar mass of BN = 5.0 g/mol (Boron) + 14.0 g/mol (Nitrogen) = 19.0 g/mol Molar mass of IF = 127.0 g/mol (Iodine) + 19.0 g/mol (Fluorine) = 146.0 g/mol Molar mass of NI_3 = 14.0 g/mol (Nitrogen) + 3 × 127.0 g/mol (Iodine) = 395.0 g/mol Molar mass of BF_3 = 5.0 g/mol (Boron) + 3 × 19.0 g/mol (Fluorine) = 67.0 g/mol

Convert grams of NI_3 to moles

Now, we can convert 30.0 g of nitrogen triiodide (NI_3) into moles using its molar mass: Moles of NI_3 = mass of NI_3 / molar mass of NI_3 Moles of NI_3 = 30.0 g / 395.0 g/mol = 0.0759 mol

Calculate the moles of IF required

Using the coefficients in the balanced equation, we can calculate the required moles of IF to produce the moles of NI_3: Moles of IF = (Moles of NI_3) × (Coefficient of IF in the balanced equation / Coefficient of NI_3 in the balanced equation) Moles of IF = 0.0759 mol × (3 / 1) = 0.2277 mol of IF

Convert moles of IF required to mass

Finally, we can find the mass of iodine monofluoride (IF) required by converting the moles back to grams using its molar mass: Mass of IF = Moles of IF × Molar mass of IF Mass of IF = 0.2277 mol × 146.0 g/mol ≈ 33.23 g Therefore, around 33.23 g of iodine monofluoride must be used to produce 30.0 g of nitrogen triiodide.

Calculate the mass of BF_3 produced

Now, we can determine the mass of by-product (BF_3) produced when 30.0 g of nitrogen triiodide is created. First, we need to find the moles of BF_3 produced, using the coefficients in the balanced equation: Moles of BF_3 = (Moles of NI_3) × (Coefficient of BF_3 / Coefficient of NI_3) Moles of BF_3 = 0.0759 mol × (1 / 1) = 0.0759 mol Next, we convert moles of BF_3 produced back into grams using its molar mass: Mass of BF_3 = Moles of BF_3 × Molar mass of BF_3 Mass of BF_3 = 0.0759 mol × 67.0 g/mol ≈ 5.09 g The maximum mass of by-product (BF_3) created when 30.0 g of nitrogen triiodide is produced is approximately 5.09 g.

Key concepts

Understanding Molar Mass

Molar mass is the concept that links the mass of a substance to its amount in moles. It's calculated by summing the atomic masses of each element in a compound. For example, to find the molar mass of Boron Nitride (BN), you add the atomic masses of boron (5.0 g/mol) and nitrogen (14.0 g/mol), giving 19.0 g/mol.
For iodine monofluoride (IF), sum the atomic mass of iodine (127.0 g/mol) and fluorine (19.0 g/mol), resulting in a molar mass of 146.0 g/mol. Nitrogen triiodide (NI₃) combines nitrogen (14.0 g/mol) with three iodines (3 x 127.0 g/mol), totaling 395.0 g/mol. Finally, boron trifluoride (BF₃) includes boron (5.0 g/mol) and three fluorines (3 x 19.0 g/mol), amounting to 67.0 g/mol.
Understanding molar mass is essential for converting between grams and moles, as seen in the exercise where mass is converted to moles and vice versa.

Exploring Chemical Reactions

A chemical reaction involves the transformation of reactants into products. In our example, boron nitride (BN) reacts with iodine monofluoride (IF) to form nitrogen triiodide (NI₃) and boron trifluoride (BF₃). Each substance behaves predictably based on who it reacts with and what it becomes.
Chemical reactions involve breaking bonds between atoms in reactants and forming new bonds to create products. These reactions must follow the law of conservation of mass, meaning that the mass and the number of atoms you start with must equal what you end up with. This fundamental rule is why we need balanced chemical equations.
By understanding the way substances change during reactions, we can predict the outcomes and determine the amounts of substances needed or produced.

The Importance of Balanced Equations

Balanced chemical equations are crucial in ensuring that reactions conform to the law of conservation of mass. This means that the number of atoms of each element in the reactants equals those in the products.
In the given reaction, the equation \[ \mathrm{BN} + 3 \mathrm{IF} \rightarrow \mathrm{NI}_3 + \mathrm{BF}_3 \] shows the balanced form. Each element has equivalent atoms on both sides: one boron, one nitrogen, three iodine, and three fluorine. This balance is necessary to accurately perform stoichiometric calculations.
If we didn’t balance the equation, calculations like determining how many grams of iodine monofluoride are required or how much of the by-product BF₃ is produced would give incorrect results. Stoichiometry relies on interpreting these balanced equations to relate quantities of reactants and products effectively.