Question

One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ $$ \mathrm{Ba}(\mathrm{SCN})_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NH}_{3}(g) $$ 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 (NH_sCN) must be used if it is to react completely with 6.5 g barium hydroxide octahydrate?

Short answer

The balanced equation for the reaction is: \(Ba(OH)_2\cdot 8H_2O(s) + 2NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + 8H_2O(l) + 2NH_3(g)\). To react completely with 6.5 g of barium hydroxide octahydrate (\(Ba(OH)_2\cdot 8H_2O\)), 3.135 g of ammonium thiocyanate (\(NH_4SCN\)) is needed.

Step-by-step solution

Balance the chemical equation

To balance the chemical equation, first, count the number of each type of atoms on both sides of the equation. Then, balance one element at a time by changing the coefficients (whole-number multipliers) of compounds until both sides have the same number of each type of atoms. Initial equation: \(Ba(OH)_2\cdot 8H_2O(s) + NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g)\) Number of atoms on the left side: Ba = 1, O = 9, H = 18, N = 1, S = 1, C = 1 Number of atoms on the right side: Ba = 1, O = 1, H = 2, N = 1, S = 2, C = 2 Add coefficient 8 in front of \(H_2O(l)\) and 2 in front of \(NH_4SCN(s)\) to balance the atoms: \(Ba(OH)_2\cdot 8H_2O(s) + 2NH_4SCN(s) \longrightarrow Ba(SCN)_2(s) + 8H_2O(l) + 2NH_3(g)\) Now, the equation is balanced.

Find the molar mass of the compounds

To find the mass of ammonium thiocyanate required, we need the molar mass of barium hydroxide octahydrate and ammonium thiocyanate. To find the molar mass, we will add the molar masses of the elements of each compound. Molar mass of barium hydroxide octahydrate, \(Ba(OH)_2\cdot 8H_2O\): Ba = 137.33 g/mol, O = 16.00 g/mol, H = 1.01 g/mol \(137.33 + 2(1.01) + 2(16.00) + 8 (2(1.01) + 16.00)= 315.51\, g/mol\) Molar mass of ammonium thiocyanate, \(NH_4SCN\): N = 14.01 g/mol, H = 1.01 g/mol, S = 32.07 g/mol, C = 12.01 g/mol \(14.01 + 4(1.01) + 32.07 + 12.01 = 76.12\, g/mol\)

Find the mass of ammonium thiocyanate needed

Now, we will use the stoichiometric ratio between barium hydroxide octahydrate and ammonium thiocyanate, and the mass of barium hydroxide octahydrate given to find the mass of ammonium thiocyanate required. From the balanced equation, the stoichiometric ratio is: \[1\:Ba(OH)_2\cdot 8H_2O : 2 NH_4SCN\] Given mass of barium hydroxide octahydrate = 6.5 g Convert this mass into moles by dividing the given mass by the molar mass: \[moles\:of\:Ba(OH)_2\cdot 8H_2O = \frac{6.5\, g}{315.51 \,g/mol} = 0.0206\, mol\] Using the stoichiometric ratio (1:2), calculate the moles of ammonium thiocyanate needed: \[moles\:of\:NH_4SCN = 0.0206\, mol\cdot 2 = 0.0412\, mol\] Convert the moles of ammonium thiocyanate into mass by multiplying by its molar mass: \[mass\:of\:NH_4SCN = 0.0412\, mol\cdot 76.12\, g/mol = 3.135\, g\] So, 3.135 g of ammonium thiocyanate is needed to react completely with 6.5 g of barium hydroxide octahydrate.

Key concepts

Balancing Chemical Equations

In chemistry, balancing chemical equations is crucial. It ensures that the same number of atoms for each element is present on both sides of the equation. This reflects the law of conservation of mass. To balance an equation, you need to adjust the coefficients, which are the numbers in front of molecules, without changing the actual compounds.

Let's consider an example involving barium hydroxide octahydrate and ammonium thiocyanate:

• Initial equation: \( Ba(OH)_2\cdot 8H_2O(s) + NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g) \)

Start by counting the number of each atom on both sides. You'll notice that some are not balanced. By adding coefficients:

• Use "2" in front of \( NH_4SCN(s) \) and "8" in front of \( H_2O(l) \).

After this, every type of atom, such as oxygen and nitrogen, is balanced on both sides, and the equation reads:

• \( Ba(OH)_2\cdot 8H_2O(s) + 2NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + 8H_2O(l) + 2NH_3(g) \)

This balanced equation respects mass conservation, indicating a properly arranged chemical reaction.

Molar Mass

Molar mass is a fundamental concept in chemistry. It determines how much one mole of a substance weighs, using grams per mole (g/mol) as its unit. To find the molar mass of a compound, add up the atomic masses of all atoms in its chemical formula.In the example of barium hydroxide octahydrate \(Ba(OH)_2 \cdot 8H_2O\):

• Ba: 137.33 g/mol
• O (9 atoms): \( 9 \times 16.00 \,g/mol \)
• H (18 atoms): \( 18 \times 1.01 \,g/mol \)

Adding these gives a molar mass of approximately 315.51 g/mol.

Similarly, for ammonium thiocyanate \(NH_4SCN\):

• N: 14.01 g/mol
• H (4 atoms): \(4 \times 1.01 \,g/mol \)
• S: 32.07 g/mol
• C: 12.01 g/mol

Summing these values gives around 76.12 g/mol for its molar mass.

This calculation aids in stoichiometric computations, helping measure reactants and products accurately.

Stoichiometric Calculations

Stoichiometry involves quantitative relationships between reactants and products in a chemical reaction. It's essentially the arithmetic of chemistry. With a balanced equation, you can use stoichiometry to determine the required reactant amounts or the amounts of products formed.For example, in the chemical reaction between barium hydroxide octahydrate and ammonium thiocyanate, the balanced equation gives us:

• \( 1 \: Ba(OH)_2\cdot 8H_2O \: \rightarrow 2 \: NH_4SCN \)

Given the mass of barium hydroxide octahydrate as 6.5 g, first find moles:

• \( moles \: of \: Ba(OH)_2\cdot 8H_2O = \frac{6.5 \, g}{315.51 \,g/mol} \approx 0.0206 \, mol \)

Using the 1:2 stoichiometric ratio from the balanced equation, calculate moles of ammonium thiocyanate needed:

• \( moles \: of \, NH_4SCN = 0.0206 \, mol \times 2 = 0.0412 \, mol \)

The final step is to convert these moles into grams:

• \( mass \: of \: NH_4SCN = 0.0412 \, mol \times 76.12 \, g/mol = 3.135 \, g \)

This calculation ensures that the correct amount of ammonium thiocyanate will react with 6.5 g of barium hydroxide octahydrate, showcasing stoichiometry's practical application.