Question

A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

Short answer

The formula of the scandium chloride produced in the reaction is ScCl₃. This is based on the balanced chemical equation and the stoichiometry, which shows a 1:3 ratio of scandium to chlorine atoms.

Step-by-step solution

Determine the amount (moles) of scandium and hydrogen involved in the reaction.

To do this, we will use their respective molar masses. The molar mass of scandium (Sc) is approximately 44.96 g/mol, and the molar mass of hydrogen (H) is approximately 1.008 g/mol. For scandium: Moles of Sc = (mass of Sc) / (molar mass of Sc) = (2.25 g) / (44.96 g/mol) = 0.05002 mol For hydrogen gas: Since hydrogen gas is H2, its molar mass = 2 × 1.008 g/mol = 2.016 g/mol Moles of H2 = (mass of H2) / (molar mass of H2) = (0.1502 g) / (2.016 g/mol) = 0.07457 mol

Obtain the balanced chemical equation for the reaction.

We know that scandium metal reacts with hydrochloric acid to produce scandium chloride and hydrogen gas. In order to balance the chemical equation, we must ensure that the number of atoms of each element on the reactants side is equal to the number of the same element on the products side. Sc(s) + nHCl(aq) → ScClₓ(aq) + (n/2)H2(g) Here, "n" refers to the number of moles of chlorine atoms reacting with one mole of scandium and "x" refers to the number of moles of chlorine combined with one mole of scandium in product.

Use stoichiometry to find the ratios of moles of elements.

We can now use stoichiometry to determine the number of moles of chlorine atoms reacting with scandium. From the balanced equation in step 2, we have: 1 mol Sc reacts with n mol Cl to produce (n/2) mol H2 According to the mole calculation in step 1, we have: 0.05002 mol Sc reacts with n' mol Cl to produce 0.07457 mol H2 From the balanced equation, n' mol Cl reacts with 0.05002 mol Sc to produce (n'/2) mol H2 Hence, we can write: (n'/2) = 0.07457 n' = 0.07457 * 2 = 0.14914 mol Since both scandium and chlorine are present in a 1:1 ratio (from the balanced equation), we can determine the ratio of moles of Sc:Cl as: Ratio (Sc:Cl) = Moles of Sc : Moles of Cl = 0.05002 : 0.14914 ≈ 1:3

Determine the formula of the scandium chloride.

Based on the ratio we found in step 3 (1:3), the formula of the scandium chloride produced would be ScCl₃. So, the formula of scandium chloride produced in the reaction is ScCl₃.

Key concepts

Molar Mass

Understanding the concept of molar mass is a fundamental aspect of stoichiometry, and indeed all of chemistry. The molar mass is defined as the mass of one mole of a substance, and it is expressed in grams per mole (g/mol). A mole is a basic unit in chemistry that represents a certain number of particles - usually atoms or molecules - and that number is Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles.

For example, in the given exercise, the molar mass of scandium (Sc) is approximately 44.96 g/mol, which means one mole of scandium atoms weighs 44.96 grams. The same concept applies to hydrogen (H). Since we usually encounter hydrogen as the diatomic molecule H₂, we calculate the molar mass as 2 times the molar mass of a single hydrogen atom, resulting in 2.016 g/mol for hydrogen gas.

When dealing with chemical reactions, calculating the molar mass allows us to convert between the mass of a substance and the number of moles, enabling us to relate the masses of reactants and products to their amounts in moles, which is critical for quantitative analysis in chemistry.

Why Understanding Molar Mass is Essential

Grasping the concept of molar mass is crucial because it serves as the bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure and observe. With molar mass, students can translate a chemical formula into a tangible quantity that can be measured in the laboratory.

Balanced Chemical Equation

The cornerstone of virtually every stoichiometric calculation is the balanced chemical equation. It represents a chemical reaction where the number of atoms for each element is the same on both the reactant and product sides, adhering to the law of conservation of mass. This balance is crucial because it shows us the exact proportions in which chemicals react.

For instance, the exercise involves a reaction between scandium metal and hydrochloric acid. The balanced chemical equation must reflect that scandium combines with chlorine from the hydrochloric acid to form scandium chloride, with hydrogen gas released in the process. Balancing this equation is a task that requires understanding both the reactants and the products, including the state of matter they are in and their formulas.

Importance of a Balanced Chemical Equation

The significance lies in its utility: without a balanced equation, it is impossible to accurately perform stoichiometric calculations, which are vital for predicting the quantities of reactants needed and products formed. Moreover, it is a visual representation that allows chemists to ensure that reactions adhere to the principles of matter conservation.

Mole Ratio

The mole ratio is simply the ratio of moles of one substance to the moles of another substance as indicated by the coefficients in a balanced chemical equation. This ratio is a key aspect in stoichiometry since it allows for the conversion of moles of one substance into moles of another.

In our exercise, calculating the mole ratio between scandium and hydrogen gas provides us with the information needed to deduce the formula of the scandium chloride produced. By understanding that chemical reactions occur according to defined mole ratios, students can calculate the amounts of reactants required or products that can be expected from a given quantity of reactant.

Applying Mole Ratios in Calculations

The effectiveness of the mole ratio lies in its ability to simplify the relationship between reactants and products in a balanced equation. It is an indispensable tool in stoichiometry for figuring out how much of one substance will react with a given amount of another substance, or for predicting the yield of a reaction. Grasping this concept allows for the quantitative analysis of chemical reactions, making it essential for students to learn and comprehend the process well.