Question

Consider the unbalanced chemical equation: $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \rightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ Suppose a \(32.9-\mathrm{g}\) sample of \(\mathrm{CaSiO}_{3}\) is reacted with \(31.8 \mathrm{~L}\) of \(\mathrm{HF}\) at \(27.0 \mathrm{C}\) and 1.00 atm. Assuming the reaction goes to completion, calculate the mass of \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

Short answer

In this reaction, 29.47 g of SiF4 and 10.20 g of H2O are produced.

Step-by-step solution

Balance the chemical equation

The given unbalanced equation is: \( CaSiO_3 (s) + HF (g) \rightarrow CaF_2 (aq) + SiF_4 (g) + H_2O (l) \) To balance the equation, we follow these steps: 1. Start with balancing elements that appear in just one compound in each side of the reaction. 2. Balance other elements subsequently. Balancing the equation: 1. Add a coefficient of 2 in front of HF. \( CaSiO_3 (s) + 2 HF (g) \rightarrow CaF_2 (aq) + SiF_4 (g) + H_2O (l) \) 2. Add a coefficient of 2 in front of the H2O. \( CaSiO_3 (s) + 2 HF (g) \rightarrow CaF_2 (aq) + SiF_4 (g) + 2 H_2O (l) \) Now, the balanced equation is: \( CaSiO_3 + 2 HF \rightarrow CaF_2 + SiF_4 + 2 H_2O \)

Identify the limiting reagent

First, we need to find the moles of each reactant. We are given 32.9 g of CaSiO3 and 31.8 L of HF. Use the molar mass of each compound to convert mass to moles for CaSiO3 and use the ideal gas law to find moles for HF. For CaSiO3: Molar mass = \(40.08 + 28.09 + (3·16.00) \,g/mol = 116.17 \,g/mol \) Moles of CaSiO3 = \( \dfrac{32.9 \,g}{116.17 \,g/mol} = 0.2832 \,mol \) For HF: Use the Ideal Gas Law: \(PV=nRT\) In this case, P = 1 atm, V = 31.8 L, T = 27°C = 300.15 K, R = 0.0821 L·atm/mol·K Moles of HF = \( \dfrac{PV}{RT} = \dfrac{(1 \,atm)(31.8 \,L)}{(0.0821 \,L·atm/mol·K)(300.15 \,K)} = 1.2889 \,mol \) Now, compare the mole ratios of the reactants to the stoichiometric ratio in the balanced equation: \(\dfrac{0.2832 \,mol \, CaSiO_3}{1} < \dfrac{1.2889 \,mol \, HF}{2}\) Thus, CaSiO3 is the limiting reagent.

Calculate the mass of SiF4 and H2O produced

Using stoichiometry with the balanced equation and the moles of the limiting reactant (CaSiO3), we can calculate the moles of SiF4 and H2O produced. Moles of SiF4 produced = (0.2832 mol CaSiO3)(1 mol SiF4/1 mol CaSiO3) = 0.2832 mol SiF4 Moles of H2O produced = (0.2832 mol CaSiO3)(2 mol H2O/1 mol CaSiO3) = 0.5664 mol H2O Now, convert the moles of SiF4 and H2O to grams by multiplying by their respective molar masses: Mass of SiF4 produced = (0.2832 mol SiF4)(104.12 g/mol) = 29.47 g SiF4 Mass of H2O produced = (0.5664 mol H2O)(18.015 g/mol) = 10.20 g H2O So, in this reaction, 29.47 g of SiF4 and 10.20 g of H2O are produced.

Key concepts

Chemical Equation Balancing

The process of chemical equation balancing is crucial to understanding stoichiometry, which involves making sure that the number of atoms for each element is equal on both sides of the equation. Balance comes from the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.

When balancing a chemical equation, one typically starts with the elements that appear in only one reactant and one product, as they are often the easiest to balance. Next, balance the remaining elements, possibly adjusting coefficients to ensure that the same number of atoms for each element is present on both sides of the equation.

Tips to Improve Understanding

• Write down the number of atoms for each element before and after the reaction to visualize the imbalance.
• Understand that coefficients in chemical equations represent the number of moles of each substance involved in the reaction.
• Practice with different types of reactions because the strategy to balance each may vary slightly.

Without a balanced equation, it's nearly impossible to progress correctly through stoichiometry problems.

Limiting Reagent

The concept of the limiting reagent is critical in stoichiometry as it determines the amount of product that can be formed in a reaction. The limiting reagent is the reactant that will be used up first, limiting the extent of the reaction.

To identify the limiting reagent, calculate the number of moles of each reactant and compare them based on the balanced chemical equation. The reactant that provides the lesser amount of product, as predicted by the stoichiometric coefficients, is the limiting reagent.

Effective Learning Strategies

• Always start with a balanced equation.
• Convert all given quantities to moles for an accurate comparison.
• Understand the stoichiometric ratios of the reactants and products.

Grasping the concept of the limiting reagent is key to predicting the amounts of products and the completion of reactions.

Mole-to-Mass Conversion

Mole-to-mass conversion is a fundamental skill in solving stoichiometry problems. To convert moles to mass, use the molar mass of the substance, which is the mass of one mole of that substance. The molar mass is usually given in grams per mole (g/mol) and is calculated by summing the atomic masses of all atoms in the molecule.

After using stoichiometry to find the number of moles of a product or reactant, multiply that number by the substance's molar mass to determine the mass in grams.

Strategies to Enhance Comprehension

• Memorize the molar masses of common elements and compounds.
• Use a periodic table to accurately calculate the molar masses for substances.
• Work on practice problems to gain confidence in moving between moles and grams.

Understanding mole-to-mass conversion is essential because measurements in the lab are often made by mass, not by counting individual atoms or molecules.