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

Step by step solution

01

Balance the chemical equation

The given unbalanced chemical equation is: \( Ba(OH)_2 \cdot 8H_2O(s) + NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g) \) To balance the equation, we will adjust the coefficients to make the number of atoms of each element the same on both sides: \( Ba(OH)_2 \cdot 8H_2O(s) + 2NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + 10H_2O(l) + 2NH_3(g) \) Now the chemical equation is balanced.

02

Find the molar masses of barium hydroxide octahydrate and ammonium thiocyanate

To perform stoichiometric calculations, we require the molar masses of the reactants. We can find these by adding the molar masses of the constituent elements in each compound: \( Molar\ mass\ of\ Ba(OH)_2 \cdot 8H_2O = 137.33 + 2(15.999) + 2(1.007) + 8(2(1.007) + 15.999) \approx 315.51\ g/mol \) \( Molar\ mass\ of\ NH_4SCN = 14.0067 + 4(1.007) + 32.06 + 12.01 + 14.0067 \approx 76.12 g/mol \)

03

Use the stoichiometry of the balanced equation to find the required mass of ammonium thiocyanate

From the balanced equation, 2 moles of ammonium thiocyanate (NH_4SCN) are required to react with 1 mole of barium hydroxide octahydrate (Ba(OH)_2·8H_2O). Given mass of barium hydroxide octahydrate = 6.5 g First, we calculate the moles of barium hydroxide octahydrate: \( Moles\ of\ Ba(OH)_2 \cdot 8H_2O = \frac{6.5 g}{315.51 g/mol} = 0.0206 mol \) Now, finding the moles of ammonium thiocyanate needed: \( Moles\ of\ NH_4SCN = 2\times Moles\ of\ Ba(OH)_2 \cdot 8H_2O = 2 \times 0.0206 mol = 0.0412 mol \) Finally, finding the mass of ammonium thiocyanate needed: \( Mass\ of\ NH_4SCN = Moles \times Molar\ Mass = 0.0412 mol \times 76.12 g/mol = 3.135 g \)

04

State the answer

The mass of ammonium thiocyanate (NH_4SCN) required to completely react with 6.5 g of barium hydroxide octahydrate (Ba(OH)_2·8H_2O) is approximately 3.135 g.

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