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?

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.

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 \(14.01 + 4(1.01) + 32.07 + 12.01 = 76.12\, g/mol\)

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.