Question

A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce \(0.1502 \mathrm{~g}\) 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.

Step-by-step solution

Identify the balanced chemical equation for the reaction

The reaction between scandium (Sc) and hydrochloric acid (HCl) can be written as: \(Sc + xHCl -> ScCl_x + x/2 H_2\) As we don't know the value of x yet, we will find the number of moles of Sc and H₂ to determine the value of x.

Calculate the moles of scandium

Using the given mass of scandium (2.25 g), we can calculate the moles of scandium by dividing this mass by the molar mass of scandium (44.96 g/mol): Moles of Sc = \(\frac{2.25 g}{44.96 g/mol}\) = 0.05 mol

Calculate the moles of hydrogen gas

Using the given mass of hydrogen gas (0.1502 g), we can calculate the moles of hydrogen gas by dividing this mass by the molar mass of hydrogen gas (2.02 g/mol): Moles of H₂ = \(\frac{0.1502 g}{2.02 g/mol}\) = 0.0743 mol Now, since there are x/2 moles of H₂ resulting from x moles of HCl: Moles of HCl = 2 * Moles of H₂ = 2 * 0.0743 mol = 0.1486 mol

Determine the moles of chlorine

Since the moles of HCl equal the moles of chlorine, we can directly calculate the moles of chlorine as: Moles of Cl = 0.1486 mol

Determine the empirical formula of scandium chloride

Now we will find the ratio of moles of Sc to moles of Cl. Divide the moles of both elements by the smaller value, in this case, the moles of Sc (0.05 mol): Ratio of Sc:Cl = \(\frac{0.05}{0.05}\):\(\frac{0.1486}{0.05}\) ≈ 1:3 Thus, the empirical formula of scandium chloride is ScCl₃, and the formula of the compound produced in the reaction is ScCl₃.

Key concepts

Chemical Reaction with Hydrochloric Acid

Understanding how metals react with hydrochloric acid is a fundamental concept in chemistry. When a metal like scandium (Sc) is introduced to hydrochloric acid (HCl), a displacement reaction occurs. This chemical process involves the scandium atoms replacing the hydrogen in hydrochloric acid, producing hydrogen gas ((H_2)) and scandium chloride (ScCl_x).

This reaction is an example of a single replacement reaction where the metal takes the place of hydrogen in the acid. Such reactions are common when studying the reactivity of metals, and they are essential for comprehending how different elements interact to form new compounds.

Mole Concept in Chemistry

The mole concept is a bridge between the atomic world and the laboratory world. It's a counting unit used in chemistry to express amounts of a chemical substance. One mole corresponds to the Avogadro number of particles, which is 6.022x10²³. Moles allow chemists to quantify substances based on the number of atoms, molecules, ions, or other entities present.

In the step-by-step solution provided for scandium chloride, the mole concept is utilized to calculate the amount of substance from a given mass. This is done by dividing the mass of the element by its molar mass (atomic weight in g/mol). This calculation is pivotal in determining the ratio of the elements in the compound, leading us to the empirical formula.

Stoichiometry Calculations

Stoichiometry calculations are centered around the quantitative relationships within a chemical reaction. They hinge on the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, the amount of reactants used and products formed are stoichiometrically related.

In our example, after determining the moles of scandium and hydrogen gas, stoichiometry plays a critical role in finding the moles of hydrochloric acid reacted, and consequently, the moles of chlorine. By using the identified moles of each reactant and product, stoichiometric calculations lead us to the empirical formula of scandium chloride.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is met. It involves making sure that the number of atoms for each element is the same on both the reactant and product sides of a chemical equation.

The exercise provides the skeleton equation for the reaction between scandium and hydrochloric acid. This equation must be balanced to reflect the accurate stoichiometry of the reaction. Here, the emphasis is on determining the subscript 'x' that will balance the equation, implying how many moles of HCl react with scandium to yield the observed moles of hydrogen gas. Once the equation is balanced, it can be used for stoichiometric calculations to predict the amounts of reactants consumed and products formed.