Problem 108 - Chemistry An Atoms First Approach

Question

One of relatively few reactions that takes place directly between two solids at room temperature is $$\begin{aligned}\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) & \longrightarrow \\\\\mathrm{Ba}(\mathrm{SCN})_{2}(s) &+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g)\end{aligned}$$ In this equation, the \(\cdot 8 \mathrm{H}_{2} \mathrm{O}\) in \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) indicates the presence of eight water molecules. This compound is called barium hydroxide octahydrate. a. Balance the equation. b. What mass of ammonium thiocyanate \(\left(\mathrm{NH}_{4} \mathrm{SCN}\right)\) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

Step-by-Step Solution

Answer

The balanced chemical equation is: \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) + 2 \mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(s) + 9 \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{NH}_{3}(g)\). To react completely with 6.5 g of barium hydroxide octahydrate, 3.138 g of ammonium thiocyanate is required.

Step 1: Balancing the chemical equation

The given chemical equation is: \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) + \mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(s) + \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{NH}_{3}(g)\) To balance this equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. 1. Balance barium (Ba) atoms: There is only 1 Ba atom on both sides, so we don't need to change any coefficients for Ba. 2. Balance oxygen (O) atoms: The right side has 9 oxygen atoms (1 in H2O and 8 in Ba(OH)2 * 8H2O). Add a coefficient of 9 in front of H2O on the left side: \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) + \mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(s) + 9 \mathrm{H}_{2} \mathrm{O}(l) + \mathrm{NH}_{3}(g)\) 3. Balance hydrogen (H) atoms: There are now 26 hydrogen atoms on the right side (2 in each of the 9 H2O molecules plus 8 in Ba(OH)2 * 8H2O). To balance hydrogen atoms, add a coefficient of 2 in front of NH3 on the left side: \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) + \mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(s) + 9 \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{NH}_{3}(g)\) 4. Balance nitrogen (N) and sulfur (S) atoms: There are now 2 nitrogen atoms on the right side (2 in 2NH3). To balance nitrogen atoms, add a coefficient of 2 in front of NH4SCN on the left side: \(\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s) + 2 \mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(s) + 9 \mathrm{H}_{2} \mathrm{O}(l) + 2 \mathrm{NH}_{3}(g)\) Now, the chemical equation is balanced.

Step 2: Finding the mass of ammonium thiocyanate

We need to find the mass of ammonium thiocyanate (NH4SCN) that reacts completely with 6.5 g of barium hydroxide octahydrate [Ba(OH)2 * 8H2O]. First, we need to calculate the molar masses of both compounds: Molar mass of Ba(OH)2 * 8H2O = 137.3 (Ba) + 2 * 15.999 (O) + 2 * 1.008 (H) + 8 * (2 * 1.008 + 15.999) = 315.5 g/mol Molar mass of NH4SCN = 1.008 * 5 + 14.007 + 12.011 + 15.999 * 2 = 76.123 g/mol Next, convert the mass of Ba(OH)2 * 8H2O to moles: 6.5 g Ba(OH)2 * 8H2O * (1 mol / 315.5 g) = 0.0206 mol Using stoichiometry from the balanced equation, we know that 1 mol of Ba(OH)2 * 8H2O reacts with 2 mol of NH4SCN. Now convert the moles of Ba(OH)2 * 8H2O to moles of NH4SCN: 0.0206 mol Ba(OH)2 * 8H2O * (2 mol NH4SCN / 1 mol Ba(OH)2 * 8H2O) = 0.0412 mol NH4SCN Finally, convert moles of NH4SCN to mass: 0.0412 mol NH4SCN * (76.123 g / mol) = 3.138 g Thus, 3.138 g of ammonium thiocyanate must be used to react completely with 6.5 g of barium hydroxide octahydrate.

Key Concepts

Stoichiometry

Stoichiometry is an essential concept in chemistry that helps us understand the quantitative relationships in a balanced chemical reaction. It involves using proportions derived from the balanced equation to calculate the amount of reactants and products involved in a reaction. In our original exercise, the stoichiometry of the reaction between barium hydroxide octahydrate and ammonium thiocyanate is crucial to determining the correct amounts of each substance required.
By balancing the chemical equation, we can see that one mole of barium hydroxide octahydrate reacts with two moles of ammonium thiocyanate. This tells us the ratio in which these substances react, which is vital for calculating the quantities needed for a complete reaction. Stoichiometry ensures that we use our reactants efficiently without leaving excess or causing shortages.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g/mol). It's a critical concept because it allows us to convert between mass and moles, two units commonly used in chemical calculations. To calculate the molar mass, we add the masses of all atoms in a molecule based on their atomic masses listed on the periodic table.
In our exercise, the molar mass of barium hydroxide octahydrate \(\mathrm{Ba(OH)}_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}\) is calculated by summing the atomic masses of each element in the compound: 137.3 g/mol for barium, added to twice the sum of oxygen and hydrogen for the hydroxide part, plus eight times the sum of water. Similarly, the molar mass of ammonium thiocyanate \(\mathrm{NH}_{4} \mathrm{SCN}\) is computed.
This calculation provides a basis for converting a given mass of a substance to moles, which can then be used in stoichiometry.

Solid-Solid Reactions

Most chemical reactions involve at least one liquid or gas phase, but solid-solid reactions are unique in that they occur between solid reactants at surface interfaces. These reactions are rare and often require specific conditions, such as mechanical pressure or high temperatures.
However, the given reaction between barium hydroxide octahydrate and ammonium thiocyanate is remarkable because it takes place at room temperature. This highlights a peculiar characteristic of some solid-solid reactions: they can occur without external energy input under specific circumstances, such as the right pressure or moisture content. Understanding such reactions is crucial in various fields, including material science and solid-state chemistry, as they offer insights into how different substances can interact without the need for solvents or high temperatures.

Chemical Reactions

At the core of chemistry are chemical reactions, which involve the rearrangement of atoms to form new substances. They can be identified by changes like the formation of a gas, a color change, or the emergence of a precipitate.
In the original exercise, the reaction turns solid reactants into a mix of solid, liquid, and gaseous products, demonstrating the diversity of outcomes chemical reactions can achieve. Here, barium hydroxide octahydrate reacts with ammonium thiocyanate to form barium thiocyanate, water, and ammonia gas.
This transformation exemplifies how chemical reactions can produce more than one type of product, each with its specific properties. By understanding the principles of chemical reactions, students can predict and manipulate chemical processes in the lab and industry.

Quantitative Analysis

Quantitative analysis in chemistry involves determining the amount of a given substance present in a sample or produced by a reaction. This process relies heavily on principles like stoichiometry and molar mass calculation to yield accurate results.
In the exercise at hand, quantitative analysis allows us to calculate the exact mass of ammonium thiocyanate needed to react completely with 6.5 g of barium hydroxide octahydrate. By determining how much of each reactant is transformed or how much product is formed, quantitative analysis provides vital data for both academic research and industrial applications.
Understanding quantitative analysis equips students with the skills to solve real-world problems, as it is used extensively in developing pharmaceuticals, creating new materials, and analyzing substances for environmental monitoring.