Question

For each of the following balanced equations, calculate how many grams of each product would be produced by complete reaction of \(15.0 \mathrm{g}\) of the reactant indicated in boldface. a. \(2 \mathbf{B C I}_{3}(s)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{B}(s)+6 \mathrm{HCl}(g)\) b. \(\mathbf{2} \mathbf{C} \mathbf{u}_{2} \mathbf{S}(s)+3 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{Cu}_{2} \mathrm{O}(s)+2 \mathrm{SO}_{2}(g)\) c. \(2 \mathrm{Cu}_{2} \mathrm{O}(s)+\mathbf{C} \mathbf{u}_{2} \mathbf{S}(s) \rightarrow 6 \mathrm{Cu}(s)+\mathrm{SO}_{2}(g)\) d. \(\mathrm{CaCO}_{3}(s)+\mathbf{S i O}_{2}(s) \rightarrow \mathrm{CaSiO}_{3}(s)+\mathrm{CO}_{2}(g)\)

Short answer

The short answers for each balanced equation are: a. Approximately 1.38g B(s) and 14.0g HCl(g) are produced. b. Approximately 13.5g Cu2O(s) and 6.04g SO2(g) are produced. c. Approximately 35.9g Cu(s) and 6.04g SO2(g) are produced. d. Approximately 29.0g CaSiO3(s) and 11.0g CO2(g) are produced.

Step-by-step solution

a. BCI3(s) + H2(g) → B(s) + HCl(g)#

We need to find how many grams of B(s) and HCl(g) are produced by 15.0g of BCI3(s). 1. Calculate the molar mass of BCI3: M(BCl3) = 1 * M(B) + 3 * M(Cl) = 1 * 10.81g/mol + 3 * 35.45 g/mol ≈ 117.16 g/mol. 2. Convert the mass of BCI3 to moles: moles of BCI3 = (15.0g) / (117.16g/mol) ≈ 0.128mol. 3. Use stoichiometry to find the moles of B and HCl produced: - 2 moles of BCI3 produce 2 moles of B and 6 moles of HCl. - From 0.128 mol of BCI3, we can calculate: moles of B = 0.128 mol * (2 mol B / 2 mol BCI3) = 0.128 mol B. moles of HCl = 0.128 mol * (6 mol HCl / 2 mol BCI3) = 0.384 mol HCl. 4. Calculate the masses of B and HCl formed: mass of B = moles of B * M(B) ≈ 0.128mol * 10.81g/mol ≈ 1.38g. mass of HCl = moles of HCl * M(HCl) ≈ 0.384mol * 36.46g/mol ≈ 14.0g. The masses of products are approximately 1.38g B(s) and 14.0g HCl(g).

b. Cu2S(s) + O2(g) → Cu2O(s) + SO2(g)#

We need to find how many grams of Cu2O(s) and SO2(g) are produced by 15.0g of Cu2S(s). 1. Calculate the molar mass of Cu2S: M(Cu2S) = 2 * M(Cu) + M(S) = 2 * 63.55g/mol + 32.07 g/mol ≈ 159.17 g/mol. 2. Convert the mass of Cu2S to moles: moles of Cu2S = (15.0g) / (159.17g/mol) ≈ 0.0943mol. 3. Use stoichiometry to find the moles of Cu2O and SO2 produced: - 2 moles of Cu2S produce 2 moles of Cu2O and 2 moles of SO2. - From 0.0943 mol of Cu2S, we can calculate: moles of Cu2O = 0.0943 mol * (2 mol Cu2O / 2 mol Cu2S) = 0.0943 mol Cu2O. moles of SO2 = 0.0943 mol * (2 mol SO2 / 2 mol Cu2S) = 0.0943 mol SO2. 4. Calculate the masses of Cu2O and SO2 formed: mass of Cu2O = moles of Cu2O * M(Cu2O) ≈ 0.0943mol * 143.09g/mol ≈ 13.5g. mass of SO2 = moles of SO2 * M(SO2) ≈ 0.0943mol * 64.06g/mol ≈ 6.04g. The masses of products are approximately 13.5g Cu2O(s) and 6.04g SO2(g).

c. Cu2O(s) + Cu2S(s) → Cu(s) + SO2(g)#

We need to find how many grams of Cu(s) and SO2(g) are produced by 15.0g of Cu2S(s). 1. Convert the mass of Cu2S to moles (we already have the molar mass calculated): moles of Cu2S = (15.0g) / (159.17g/mol) ≈ 0.0943mol. 2. Use stoichiometry to find the moles of Cu and SO2 produced: - 1 mole of Cu2S produces 6 moles of Cu and 1 mole of SO2. - From 0.0943 mol of Cu2S, we can calculate: moles of Cu = 0.0943 mol * (6 mol Cu / 1 mol Cu2S) = 0.5658 mol Cu. moles of SO2 = 0.0943 mol * (1 mol SO2 / 1 mol Cu2S) = 0.0943 mol SO2. 3. Calculate the masses of Cu and SO2 formed: mass of Cu = moles of Cu * M(Cu) ≈ 0.5658mol * 63.55g/mol ≈ 35.9g. mass of SO2 = moles of SO2 * M(SO2) ≈ 0.0943mol * 64.06g/mol ≈ 6.04g. The masses of products are approximately 35.9g Cu(s) and 6.04g SO2(g).

d. CaCO3(s) + SiO2(s) → CaSiO3(s) + CO2(g)#

We need to find how many grams of CaSiO3(s) and CO2(g) are produced by 15.0g of SiO2(s). 1. Calculate the molar mass of SiO2: M(SiO2) = M(Si) + 2 * M(O) = 28.09g/mol + 2 * 16.00 g/mol ≈ 60.09 g/mol. 2. Convert the mass of SiO2 to moles: moles of SiO2 = (15.0g) / (60.09g/mol) ≈ 0.2498mol. 3. Use stoichiometry to find the moles of CaSiO3 and CO2 produced: - 1 mole of SiO2 produces 1 mole of CaSiO3 and 1 mole of CO2. - From 0.2498 mol of SiO2, we can calculate: moles of CaSiO3 = 0.2498 mol * (1 mol CaSiO3 / 1 mol SiO2) = 0.2498 mol CaSiO3. moles of CO2 = 0.2498 mol * (1 mol CO2 / 1 mol SiO2) = 0.2498 mol CO2. 4. Calculate the masses of CaSiO3 and CO2 formed: mass of CaSiO3 = moles of CaSiO3 * M(CaSiO3) ≈ 0.2498mol * 116.16g/mol ≈ 29.0g. mass of CO2 = moles of CO2 * M(CO2) ≈ 0.2498mol * 44.01g/mol ≈ 11.0g. The masses of products are approximately 29.0g CaSiO3(s) and 11.0g CO2(g).

Key concepts

Chemical Reactions

In chemistry, a chemical reaction is a process where reactants transform into products. The reaction involves breaking bonds in the reactants and forming new bonds to create the products. Balance is crucial in chemical reactions, meaning the number of atoms for each element must be equal on both sides of the equation. This is called the conservation of matter. For instance, when BCl₃ reacts with H₂, it transforms into B and HCl. The chemical equation for this reaction must show equal numbers of each type of atom on both sides to be balanced. This concept ensures that no atoms are lost or gained in the reaction.
A balanced chemical reaction helps predict the amounts of products formed from given quantities of reactants. This is where stoichiometry comes into play, allowing chemists to calculate precisely how much product can be produced. Understanding this principle is essential for performing experiments efficiently and optimizing chemical processes.

Mole Concept

The mole is a fundamental concept in chemistry for quantifying amounts of a substance. It represents Avogadro's number (6.022 × 10²³) of particles, such as atoms, molecules, or ions. The mole allows chemists to count these extremely small entities in manageable figures. For practical calculations, it connects mass to the number of particles through the molar mass.
For instance, when working with the reaction of BCl₃ with H₂, understanding the mole concept allows us to determine how many moles of products (B and HCl) are formed from a given mass of BCl₃. By converting the mass of BCl₃ (15.0g) to moles using its molar mass, we know the exact number of molecules involved.
This approach is useful in laboratories and industrial settings where precise measurements of chemicals are required. It ensures accuracy in scientific studies and experimentation.

Molar Mass

Molar mass is the weight of one mole of a substance, usually expressed in grams per mole (g/mol). It is a critical factor for converting between grams and moles in chemical calculations. Each element has a specific atomic mass, and the molar mass of a compound is the sum of the atomic masses of its constituent atoms.
To find the molar mass of BCl₃, we add the molar mass of boron with three times the molar mass of chlorine: \( M(BCl₃) = 1 \times 10.81 \text{g/mol} + 3 \times 35.45 \text{g/mol} \approx 117.16 \text{g/mol}\). This calculation allows us to convert the mass of BCl₃ to moles, setting the stage for using stoichiometry in determining the mass of products.
Understanding molar mass is vital for getting accurate results in predictive calculations and essential for interpreting data in real-world chemical processes. It simplifies the relationship between the amount of substance and its mass, providing a bridge between the atomistic world and measurable quantities.