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?

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.

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.