One of relatively few reactions that takes place directly between two solids at room temperature is $$ \mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+\mathrm{NH}_{4} \mathrm{SCN}(s) \longrightarrow $$ 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 pres- ence 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?

First, we need to identify the products of the reaction. The given reactants are barium hydroxide octahydrate (Ba(OH)2 · 8H2O) and ammonium thiocyanate (NH4SCN). The products of this reaction are barium thiocyanate (Ba(SCN)2) and water (H2O) and ammonia (NH3). Unbalanced equation: \(Ba(OH)_2 \cdot 8 H_2O(s) + NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g) \) Now, balance the elements in the equation one by one: 1. There are 2 SCN in the product side, so we need 2 moles of NH4SCN on the reactant side. Update the equation: \(Ba(OH)_2 \cdot 8 H_2O(s) + 2NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + H_2O(l) + NH_3(g) \) 2. Now, balance the hydrogen and oxygen atoms. There are (2+16) 18 hydrogen atoms and 8 oxygen atoms on the reactant side. On the product side, we require 1 mole of water and 2 moles of ammonia to balance these atoms. Update the equation: \(Ba(OH)_2 \cdot 8 H_2O(s) + 2NH_4SCN(s) \rightarrow Ba(SCN)_2(s) + 9H_2O(l) + 2NH_3(g) \) Now the chemical equation is balanced.

To calculate the mass of ammonium thiocyanate (NH4SCN) required, we can use stoichiometry. The balanced chemical equation gives us the mole ratio of reactants and products. From the balanced equation: \(1 \: mole \: Ba(OH)_2 \cdot 8H_2O \: \rightarrow \: 2 \: moles \: NH_4SCN\) First, we need to calculate the number of moles of barium hydroxide octahydrate (Ba(OH)2 · 8H2O) in 6.5 g. Molar mass of Ba(OH)2 · 8H2O: \(137.3 (Ba) + 2 \times 15.999 (O) + 2 \times 1.008 (H) + 8 \times (2 \times 1.008 (H) + 15.999 (O)) \\ = 137.3 + 31.998 + 16.032 + 8 \times 18.015 \\ = 315.51 \: g/mol\) Moles of Ba(OH)2 · 8H2O = \(\frac{6.5}{315.51} = 0.0206 \: moles\) Using stoichiometry, we can find the moles of ammonium thiocyanate (NH4SCN) required: Moles of NH4SCN = \(0.0206 \times 2 = 0.0412 \: moles\) Now, calculate the mass of 0.0412 moles of NH4SCN: Molar mass of NH4SCN: \(14.007 (N) + 4 \times 1.008 (H) + 32.065 (S) + 12.011 (C) + 14.007 (N) \\ = 14.007 + 4.032 + 32.065 + 12.011 + 14.007 \\ = 76.122 \: g/mol\) Mass of NH4SCN = \(0.0412 \times 76.122 = 3.14 \: g\) Thus, 3.14 g of ammonium thiocyanate (NH4SCN) must be used to react completely with 6.5 g of barium hydroxide octahydrate (Ba(OH)2 · 8H2O).